The line l has equation \(\mathbf{r} = \mathbf{i} - 2\mathbf{j} - 3\mathbf{k} + \lambda\bigl(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\bigr)\). The points A and B have position vectors \(-2\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} - \mathbf{j} + \mathbf{k}\) respectively.

(a) The direction vector of line l is \(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). The magnitude of this vector is \(\sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{6}\). Therefore, a unit vector in the direction of l is \(\dfrac{1}{\sqrt{6}}\bigl(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\bigr)\).

(b) The vector from A to B is \((3\mathbf{i} - \mathbf{j} + \mathbf{k}) - (-2\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\). Thus, a vector equation for line m is \(\mathbf{r} = -2\mathbf{i} + 2\mathbf{j} - \mathbf{k} + \mu\bigl(5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\bigr)\).

(c) The direction vector of l is \(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and for m is \(5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\). These vectors are not scalar multiples of each other, so the lines are not parallel. Solving the system obtained by equating components of the general points on l and m does not yield a consistent solution, indicating the lines do not intersect. Therefore, lines l and m are skew.

(a) To find the vector equation for the line through A and B, we first find the direction vector \(\mathbf{AB}\) by subtracting the position vector of A from B:

\(\mathbf{AB} = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -\mathbf{i} - 3\mathbf{j} + \mathbf{k}\).

The vector equation of the line through A and B is:

\(\mathbf{r} = 2\mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} - 3\mathbf{j} + \mathbf{k})\).

(b) To find the acute angle between the directions of \(AB\) and \(l\), we use the dot product formula:

\(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\).

Direction vector of \(l\) is \(\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}\).

\(\mathbf{AB} \cdot (\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}) = (-1)(1) + (-3)(-3) + (1)(-2) = -1 + 9 - 2 = 6\).

\(\|\mathbf{AB}\| = \sqrt{(-1)^2 + (-3)^2 + (1)^2} = \sqrt{11}\).\)

\(\|\mathbf{l}\| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{14}\).\)

\(\cos \theta = \frac{6}{\sqrt{11} \cdot \sqrt{14}}\).

\(\theta = \cos^{-1}\left(\frac{6}{\sqrt{11} \cdot \sqrt{14}}\right) \approx 61.1^\circ\).

(c) To show that the line through A and B does not intersect the line \(l\), express the general points of both lines in component form and equate them:

Line through A and B: \((2 - \lambda, 1 - 3\lambda, 1 + \lambda)\).

Line \(l\): \((1 + \mu, 2 - 3\mu, -3 - 2\mu)\).

Equate components and solve for \(\lambda\) and \(\mu\):

\(2 - \lambda = 1 + \mu\), \(1 - 3\lambda = 2 - 3\mu\), \(1 + \lambda = -3 - 2\mu\).

Solving these gives inconsistent results, confirming the lines do not intersect.

(a) The direction vector \(\overrightarrow{AB}\) is calculated as:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\)

The vector equation of the line is:

\(\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\)

(b) Since \(\overrightarrow{AC} = 3\overrightarrow{AB}\), we have:

\(\overrightarrow{AC} = 3 \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 3 \\ 6 \end{pmatrix}\)

Thus, \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \begin{pmatrix} -3 \\ 3 \\ 6 \end{pmatrix} = \begin{pmatrix} -2 \\ 5 \\ 5 \end{pmatrix}\)

(c) Let \(\overrightarrow{OP} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}\).

The modulus of \(\overrightarrow{OP}\) is \(\sqrt{14}\):

\(\left| \begin{pmatrix} 1 - \lambda \\ 2 + \lambda \\ -1 + 2\lambda \end{pmatrix} \right| = \sqrt{(1 - \lambda)^2 + (2 + \lambda)^2 + (-1 + 2\lambda)^2} = \sqrt{14}\)

Equating and simplifying gives:

\(3\lambda^2 - \lambda - 4 = 0\)

Solving this quadratic equation, we find \(\lambda = -1\) or \(\lambda = \frac{4}{3}\).

Thus, the position vectors are:

For \(\lambda = -1\): \(\overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}\)

For \(\lambda = \frac{4}{3}\): \(\overrightarrow{OP} = \begin{pmatrix} -\frac{1}{3} \\ \frac{10}{3} \\ \frac{5}{3} \end{pmatrix}\)

(a) To show that lines l and m are perpendicular, calculate the dot product of their direction vectors. The direction vector of l is \(4\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and for m is \(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

Calculate the dot product: \((4)(-1) + (-1)(2) + (3)(2) = -4 - 2 + 6 = 0\).

Since the dot product is zero, the lines are perpendicular.

(b) To find the intersection, equate the parametric equations of l and m:

\(3 + 4s = 1 - t\)

\(2 - s = -1 + 2t\)

\(5 + 3s = -2 + 2t\)

Solving these, we find \(s = -1\) and \(t = 2\).

Substitute back to verify all equations are satisfied. The position vector of the intersection is \(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\).

(c) To find the perpendicular distance from the origin to line m, use the formula for the distance from a point to a line in vector form. The point is the origin \(\mathbf{0}\) and the line is \(\mathbf{r} = \mathbf{i} - \mathbf{j} - 2\mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).

Using the formula, the distance \(d\) is given by:

\(d = \frac{|(-1)(-1) + (-1)(2) + (-2)(2)|}{\sqrt{(-1)^2 + 2^2 + 2^2}} = \frac{|1 - 2 - 4|}{\sqrt{9}} = \frac{5}{3}\).

Thus, the length of the perpendicular is \(\frac{1}{3}\sqrt{5}\).

(a) First, find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\mathbf{i} + 3\mathbf{j} + \mathbf{k}) - (-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\).

Given \(\overrightarrow{DC} = 3\overrightarrow{AB}\), we have:

\(\overrightarrow{DC} = 3(2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 6\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\).

Since \(\overrightarrow{DC} = \overrightarrow{OD} - \overrightarrow{OC}\), we find \(\overrightarrow{OD}\):

\(\overrightarrow{OD} = \overrightarrow{DC} + \overrightarrow{OC} = (6\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) + (2\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) = -4\mathbf{i} - \mathbf{j} + 3\mathbf{k}\).

(b) The vector equation for the line through A and B is:

\(\mathbf{r} = -\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} - 2\mathbf{k})\).

(c) To find the distance between the parallel sides, consider the perpendicular distance from C to line AB. The vector \(\overrightarrow{CP}\) is perpendicular to \(\overrightarrow{AB}\), so:

\(\overrightarrow{CP} = (3 - 2\lambda, -\lambda, -6 + 2\lambda)\).

Setting the dot product \(\overrightarrow{CP} \cdot \overrightarrow{AB} = 0\), solve for \(\lambda\):

\((3 - 2\lambda) \cdot 2 + (-\lambda) \cdot 1 + (-6 + 2\lambda) \cdot (-2) = 0\).

Simplifying gives \(\lambda = 2\).

The perpendicular distance is 3.

The area of the trapezium is given by:

\(\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} = \frac{1}{2} \times (\|\overrightarrow{AB}\| + \|\overrightarrow{DC}\|) \times 3 = 18\).

(a) To find the direction vector of \(AB\), calculate \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix}\).

The direction vector of line \(l\) is \(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\).

Calculate the dot product: \(\begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 2 \times 1 + (-1) \times (-2) + (-3) \times 1 = 2 + 2 - 3 = 1\).

Calculate the magnitudes: \(\| \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix} \| = \sqrt{2^2 + (-1)^2 + (-3)^2} = \sqrt{14}\) and \(\| \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}\).

Use the cosine formula: \(\cos \theta = \frac{1}{\sqrt{6} \sqrt{14}}\).

Find \(\theta\) using inverse cosine: \(\theta = \cos^{-1} \left( \frac{1}{\sqrt{6} \sqrt{14}} \right) \approx 83.7^\circ\) or \(1.46\) radians.

(b) Express \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) in terms of \(\lambda\): \(\overrightarrow{AP} = \begin{pmatrix} 1 + \lambda \\ 1 - 2\lambda \\ \lambda \end{pmatrix}\) and \(\overrightarrow{BP} = \begin{pmatrix} -1 + \lambda \\ 2 - 2\lambda \\ 3 + \lambda \end{pmatrix}\).

Equate the magnitudes: \((1 + \lambda)^2 + (1 - 2\lambda)^2 + \lambda^2 = (-1 + \lambda)^2 + (2 - 2\lambda)^2 + (3 + \lambda)^2\).

Simplify and solve for \(\lambda\): \(83\lambda^2 - 528\lambda + 207 = 0\).

Solving gives \(\lambda = 6\).

Substitute \(\lambda = 6\) into the line equation to find \(P\): \(\mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + 6 \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ -9 \\ 7 \end{pmatrix}\).

(a) To show that the lines are skew, express the general point of each line in component form:

Line 1: \((1 + 2s, 3 - s, 2 + 3s)\)

Line 2: \((2 + t, 1 - t, 4 + 4t)\)

Equate the components to find \(s\) and \(t\):

\(1 + 2s = 2 + t\)

\(3 - s = 1 - t\)

\(2 + 3s = 4 + 4t\)

Solving these, possible values are \(s = -1, 6, \frac{2}{5}\) and \(t = -3, 4, -\frac{1}{5}\).

Verify that all three component equations are not satisfied simultaneously, showing the lines are not parallel and thus skew.

(b) To find the acute angle between the directions of the two lines, calculate the scalar product of the direction vectors:

Direction vector of Line 1: \(\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\)

Direction vector of Line 2: \(\begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}\)

Scalar product: \(2 \times 1 + (-1) \times (-1) + 3 \times 4 = 2 + 1 + 12 = 15\)

Calculate the moduli of the direction vectors:

\(\| \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}\)

\(\| \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} \| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{18}\)

Cosine of the angle \(\theta\):

\(\cos \theta = \frac{15}{\sqrt{14} \times \sqrt{18}}\)

\(\theta = \cos^{-1}\left(\frac{15}{\sqrt{14} \times \sqrt{18}}\right) \approx 19.1^\circ\) or \(0.333\) radians.

(a) The position vector of \(M\) is \(\mathbf{OM} = 4\mathbf{i} + 2\mathbf{j}\).

For \(N\), since \(DN = 3NC\), \(N\) divides \(CD\) in the ratio 3:1. The position vector of \(C\) is \(4\mathbf{j}\) and \(D\) is \(4\mathbf{k}\). Using the section formula, \(\mathbf{ON} = \frac{1}{4}(3 \times 4\mathbf{k} + 1 \times 4\mathbf{j}) = 3\mathbf{j} + \mathbf{k}\).

The direction vector \(\mathbf{MN} = \mathbf{ON} - \mathbf{OM} = (3\mathbf{j} + \mathbf{k}) - (4\mathbf{i} + 2\mathbf{j}) = -4\mathbf{i} + \mathbf{j} + \mathbf{k}\).

The vector equation of the line through \(M\) and \(N\) is \(\mathbf{r} = 3\mathbf{j} + \mathbf{k} + \lambda(4\mathbf{i} - \mathbf{j} - \mathbf{k})\).

(b) The direction vector of \(MN\) is \(-4\mathbf{i} + \mathbf{j} + \mathbf{k}\), with magnitude \(\sqrt{(-4)^2 + 1^2 + 1^2} = \sqrt{18}\).

The projection of \(\mathbf{OM}\) on \(\mathbf{MN}\) is \(\frac{\mathbf{OM} \cdot \mathbf{MN}}{\|\mathbf{MN}\|} = \frac{(4\mathbf{i} + 2\mathbf{j}) \cdot (-4\mathbf{i} + \mathbf{j} + \mathbf{k})}{\sqrt{18}} = \frac{-16 + 2}{\sqrt{18}} = \frac{-14}{\sqrt{18}}\).

The perpendicular distance is \(\sqrt{\|\mathbf{OM}\|^2 - \left(\frac{-14}{\sqrt{18}}\right)^2} = \sqrt{20 - \frac{196}{18}} = \frac{1}{3}\sqrt{82}\).

(a) Calculate \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ -4 \end{pmatrix}.\)

Calculate \(\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}.\)

Find the magnitudes: \(AB = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{24}, \quad CD = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}.\)

Thus, \(AB = 2CD.\)

(b) The angle \(\theta\) between \(\overrightarrow{AB}\) and \(\overrightarrow{CD}\) is given by \(\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|}.\)

Calculate the dot product: \(\overrightarrow{AB} \cdot \overrightarrow{CD} = 2 \times 2 + (-2) \times 1 + (-4) \times 1 = 2.\)

Thus, \(\cos \theta = \frac{2}{\sqrt{24} \times \sqrt{6}} = \frac{2}{12} = \frac{1}{6}.\)

Therefore, \(\theta = \cos^{-1} \left( \frac{1}{6} \right) \approx 99.6^\circ.\)

(c) The vector equations of the lines are:

Line through A and B: \(\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ -4 \end{pmatrix}.\)

Line through C and D: \(\mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}.\)

Equate components and solve for \(\lambda\) and \(\mu\):

\(2 + 2\lambda = 1 + 2\mu, \quad 1 - 2\lambda = 1 + \mu, \quad 5 - 4\lambda = 2 + \mu.\)

Solving these gives inconsistent results, confirming the lines do not intersect.

(a) Express the general point of at least one line in component form: \((1 + a\lambda, 2 + 2\lambda, 1 - \lambda)\) or \((2 + 2\mu, 1 - \mu, -1 + \mu)\).

Equate at least two pairs of corresponding components and solve for \(\lambda\) or \(\mu\):

\(1 + a\lambda = 2 + 2\mu\)

\(2 + 2\lambda = 1 - \mu\)

\(1 - \lambda = -1 + \mu\)

Solving gives \(\lambda = -3\) or \(\mu = 5\).

Obtain \(a = \frac{11}{3}\).

The point of intersection has position vector \(12\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\).

(b) Use the correct process for finding the scalar product of direction vectors for the two lines: \((a, 2, -1)\) and \((2, -1, 1)\).

Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to \(\frac{1}{6}\).

State a correct equation in \(a\) in any form, e.g., \(\frac{2a - 2 - 1}{\sqrt{6(a^2 + 5)}} = \pm \frac{1}{6}\).

Solve for \(a\):

\(36(2a - 3)^2 = 6a^2 + 5\)

\(23a^2 - 72a + 49 = 0\)

Obtain \(a = 1\) or \(a = \frac{49}{23}\).

(a) The position vector of M is the midpoint of OA, which is \(\frac{1}{2}(6\mathbf{i} + 2\mathbf{j}) = 3\mathbf{i} + \mathbf{j}\).

The position vector of N is given by dividing the segment AB in the ratio 2:1. Thus, \(\overrightarrow{ON} = \frac{1}{3}(2\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{3}(2(6\mathbf{i} + 2\mathbf{j}) + (2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})) = \frac{10}{3}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

The direction vector \(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \left( \frac{10}{3}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \right) - (3\mathbf{i} + \mathbf{j}) = \frac{1}{3}\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).

Thus, the vector equation for the line through M and N is \(\mathbf{r} = 3\mathbf{i} + \mathbf{j} + \lambda \left( \frac{1}{3}\mathbf{i} + \mathbf{j} + 2\mathbf{k} \right)\).

(b) The line through O and B is \(\mathbf{r} = \mu(2\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\).

Equating the components of \(\overrightarrow{MN}\) and \(\overrightarrow{OB}\), we solve for \(\lambda\) and \(\mu\):

\(\lambda = 3\) or \(\mu = 2\).

Thus, the position vector of P is \(4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}\).

(c) To find angle OPM, calculate the scalar product of direction vectors for OP and MP:

\(\overrightarrow{OP} = 4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}\)

\(\overrightarrow{MP} = (4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}) - (3\mathbf{i} + \mathbf{j}) = \mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\)

Calculate the dot product and magnitudes:

\(\overrightarrow{OP} \cdot \overrightarrow{MP} = 4 \times 1 + 4 \times 3 + 6 \times 6 = 52\)

\(|\overrightarrow{OP}| = \sqrt{4^2 + 4^2 + 6^2} = \sqrt{68}\)

\(|\overrightarrow{MP}| = \sqrt{1^2 + 3^2 + 6^2} = \sqrt{46}\)

\(\cos \theta = \frac{52}{\sqrt{68} \times \sqrt{46}}\)

\(\theta = \cos^{-1}\left(\frac{52}{\sqrt{68} \times \sqrt{46}}\right) \approx 21.6^{\circ}\)

(a) To find the value of c, we use the fact that the angle between the lines is 60°. The direction vectors of l and m are \(\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\) and \(\begin{pmatrix} -2 \\ 4 \\ c \end{pmatrix}\) respectively. The cosine of the angle between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:

\(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\)

Substituting the given vectors and \(\theta = 60^{\circ}\), we have:

\(\cos 60^{\circ} = \frac{1(-2) + 1(4) + 2c}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{(-2)^2 + 4^2 + c^2}}\)

\(\frac{2 + 2c}{\sqrt{6} \sqrt{20 + c^2}} = \frac{1}{2}\)

Solving for c, we get:

\(2 + 2c = \frac{1}{2} \sqrt{120 + 6c^2}\)

\(4 + 4c = \sqrt{120 + 6c^2}\)

Squaring both sides:

\(16 + 16c + 16c^2 = 120 + 6c^2\)

\(10c^2 + 16c - 104 = 0\)

Solving this quadratic equation gives \(c = 2\).

(b) To find the length of the perpendicular from the point (6, -3, 6) to the line l, consider a general point on l given by \(\begin{pmatrix} 3 + \lambda \\ -2 + \lambda \\ 1 + 2\lambda \end{pmatrix}\). The vector from (6, -3, 6) to this point is:

\(\begin{pmatrix} 3 + \lambda - 6 \\ -2 + \lambda + 3 \\ 1 + 2\lambda - 6 \end{pmatrix} = \begin{pmatrix} -3 + \lambda \\ 1 + \lambda \\ -5 + 2\lambda \end{pmatrix}\)

For the perpendicular, the dot product with the direction vector of l must be zero:

\((-3 + \lambda) \cdot 1 + (1 + \lambda) \cdot 1 + (-5 + 2\lambda) \cdot 2 = 0\)

\(-3 + \lambda + 1 + \lambda - 10 + 4\lambda = 0\)

\(6\lambda - 12 = 0\)

\(\lambda = 2\)

The perpendicular vector is \(\begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix}\). Its length is:

\(\sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{11}\)

(a) To show that angle ABC is a right angle, we need to find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\).

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 5\mathbf{k}) = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{i} + \mathbf{j} + \mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = -2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\)

Calculate the scalar product: \(\overrightarrow{AB} \cdot \overrightarrow{BC} = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \cdot (-2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) = -2 - 2 + 4 = 0\)

Since the scalar product is zero, angle ABC is a right angle.

(b) To show that triangle ABC is isosceles, calculate the lengths of \(AB\) and \(BC\).

\(|\overrightarrow{AB}| = \sqrt{(1)^2 + (2)^2 + (-2)^2} = \sqrt{9} = 3\)

\(|\overrightarrow{BC}| = \sqrt{(-2)^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3\)

Since \(AB = BC\), triangle ABC is isosceles.

(c) To find the exact length of the perpendicular from O to the line through B and C, use the vector equation of the line: \(\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(-2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\).

Taking a general point \(P\) on the line, form an equation by setting the derivative of \(|\overrightarrow{OP}|\) to zero or using the scalar product method.

Solve and obtain \(\lambda = -\frac{5}{9}\).

Calculate the perpendicular distance: \(\frac{1}{3} \sqrt{2}\).

(a) To find \(\overrightarrow{OM}\), note that \(M\) divides \(AB\) in the ratio 1:2. Since \(A = 2\mathbf{i}\) and \(B = 2\mathbf{i} + 3\mathbf{j}\), \(M = 2\mathbf{i} + \frac{1}{3}(3\mathbf{j}) = 2\mathbf{i} + \mathbf{j}\). Thus, \(\overrightarrow{OM} = 2\mathbf{i} + \mathbf{j}\).

For \(\overrightarrow{MN}\), since \(N\) is the midpoint of \(FG\), \(N = 2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\). Therefore, \(\overrightarrow{MN} = (2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) - (2\mathbf{i} + \mathbf{j}) = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

(b) The vector equation of the line through \(M\) and \(N\) is \(\mathbf{r} = 2\mathbf{i} + \mathbf{j} + \lambda(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\), where \(\lambda\) is a parameter.

(c) To find the position vector of \(P\), the foot of the perpendicular from \(D\) to the line through \(M\) and \(N\), we use the fact that \(\overrightarrow{DP}\) is perpendicular to \(\overrightarrow{MN}\). Let \(\overrightarrow{DP} = (2 - \lambda)\mathbf{i} + (1 + 2\lambda)\mathbf{j} + (-2 + 2\lambda)\mathbf{k}\). The dot product \(\overrightarrow{DP} \cdot \overrightarrow{MN} = 0\) gives \(\lambda = \frac{4}{9}\). Substituting \(\lambda\) back, the position vector of \(P\) is \(\frac{14}{9}\mathbf{i} + \frac{17}{9}\mathbf{j} + \frac{8}{9}\mathbf{k}\).

To find the value of a, we express the general points of lines l and m in component form:

\(Line l: r = (a + λ)i + (2 - 2λ)j + (3 + 3λ)k\)

\(Line m: r = (2 + 2μ)i + (1 - μ)j + (2 + μ)k\)

Since the lines intersect, equate the corresponding components:

\(1. a + λ = 2 + 2μ\)

\(2. 2 - 2λ = 1 - μ\)

\(3. 3 + 3λ = 2 + μ\)

From equation (2):

\(μ = 2 - 2λ - 1 = 1 - 2λ\)

\(Substitute μ = 1 - 2λ into equation (1):\)

\(a + λ = 2 + 2(1 - 2λ)\)

\(a + λ = 2 + 2 - 4λ\)

\(a + λ = 4 - 4λ\)

\(5λ = 4 - a\)

\(λ = \frac{4 - a}{5}\)

\(Substitute λ = \frac{4 - a}{5} into equation (3):\)

\(3 + 3\left(\frac{4 - a}{5}\right) = 2 + (1 - 2\left(\frac{4 - a}{5}\right))\)

\(3 + \frac{12 - 3a}{5} = 2 + 1 - \frac{8 - 2a}{5}\)

\(3 + \frac{12 - 3a}{5} = 3 - \frac{8 - 2a}{5}\)

Multiply through by 5 to clear fractions:

\(15 + 12 - 3a = 15 - 8 + 2a\)

\(27 - 3a = 7 + 2a\)

\(5a = 20\)

\(a = -6\)

1. Find \(\overrightarrow{PQ}\) for a general point Q on l, e.g., \(-3\mathbf{i} + 6\mathbf{k} + \mu(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\).

2. Calculate the scalar product of \(\overrightarrow{PQ}\) and a direction vector for l and equate the result to zero.

3. Solve for \(\mu\) and obtain \(\mu = 2\).

4. Carry out a complete method for finding the length of \(\overrightarrow{PQ}\).

5. Obtain answer 3.

First, find the vector equation of the line passing through A and B. The direction vector AB is given by:

\(\text{Direction vector } \overrightarrow{AB} = (3\mathbf{i} + \mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\)

The vector equation of the line through A and B is:

\(\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})\)

Equate the components of the general points on both lines:

\((1 + 2\lambda, 2 - \lambda, -1 + 2\lambda) = (2 + \mu, 1 + \mu, 1 + 2\mu)\)

Solving these equations:

1. \(1 + 2\lambda = 2 + \mu\)

2. \(2 - \lambda = 1 + \mu\)

3. \(-1 + 2\lambda = 1 + 2\mu\)

From equation 1: \(\lambda = \frac{1}{2} \mu + \frac{1}{2}\)

From equation 2: \(\lambda = -\mu + 1\)

From equation 3: \(\lambda = \mu + 1\)

Substitute \(\lambda = \mu + 1\) into \(\lambda = \frac{1}{2} \mu + \frac{1}{2}\):

\(\mu + 1 = \frac{1}{2} \mu + \frac{1}{2}\)

\(\mu = -1\)

Substitute \(\mu = -1\) into \(\lambda = \mu + 1\):

\(\lambda = 0\)

Verify with the third equation:

\(-1 + 2(0) = 1 + 2(-1)\)

\(-1 \neq -1\)

Since the equations are not satisfied, the lines do not intersect.

(i) The vector equation of the line passing through A and B is \(\mathbf{r} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} + \lambda(2\mathbf{i} - 2\mathbf{k})\).

Equating the components of the line l and the line through A and B gives:

\(4 + \mu = 2 + 2\lambda\)

\(6 + 2\mu = 1\)

\(-2\mu = 3 - 3\lambda\)

Solving these equations, we find that they are inconsistent, confirming that the lines do not intersect.

(ii) The direction vector for \(\overrightarrow{AP}\) is \((2 + t, 5 + 2t, -3 - 2t)\).

Using the cosine of the angle formula, \(\cos 120^{\circ} = -\frac{1}{2}\), we set up the equation:

\(\frac{(2 + t)(2) + (5 + 2t)(0) + (-3 - 2t)(-2)}{\sqrt{(2 + t)^2 + (5 + 2t)^2 + (-3 - 2t)^2} \cdot \sqrt{4 + 0 + 4}} = -\frac{1}{2}\)

Simplifying, we obtain \(3t^2 + 8t + 4 = 0\).

Solving this quadratic equation, we find \(t = -2\) or \(t = -\frac{2}{3}\). Using \(t = -2\), the position vector of P is \(2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).

To determine if the lines are skew, we need to check if they are neither parallel nor intersecting.

First, equate the components of the lines:

For the i component: \(2 + 2s = 1 + t\)

For the j component: \(-1 + 3s = 3 + 2t\)

For the k component: \(1 - s = 4 + t\)

Solving these equations:

From \(2 + 2s = 1 + t\), we get \(t = 2s - 1\).

From \(-1 + 3s = 3 + 2t\), substitute \(t = 2s - 1\):

\(-1 + 3s = 3 + 2(2s - 1)\)

\(-1 + 3s = 3 + 4s - 2\)

\(-1 + 3s = 1 + 4s\)

\(-2 = s\)

Substitute \(s = -2\) into \(t = 2s - 1\):

\(t = 2(-2) - 1 = -4 - 1 = -5\)

Check the k component: \(1 - s = 4 + t\)

\(1 - (-2) = 4 + (-5)\)

\(3 \neq -1\)

The equations do not satisfy all components simultaneously, so the lines do not intersect.

Since the direction vectors \((2, 3, -1)\) and \((1, 2, 1)\) are not scalar multiples, the lines are not parallel.

Therefore, the lines are skew.

To find the length of the perpendicular from point \(P\) to the line \(l\), we can use the vector projection method.

1. The direction vector of the line \(l\) is \(\mathbf{d} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\).

2. The position vector of point \(P\) is \(\mathbf{p} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\).

3. A point \(A\) on the line \(l\) can be represented as \(\mathbf{a} = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}\).

4. The vector \(\mathbf{PA}\) is \(\mathbf{p} - \mathbf{a} = (3 - 4)\mathbf{i} + (-2 - 2)\mathbf{j} + (1 - 5)\mathbf{k} = -\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}\).

5. The projection of \(\mathbf{PA}\) onto \(\mathbf{d}\) is given by:

\(\text{proj}_{\mathbf{d}} \mathbf{PA} = \frac{\mathbf{PA} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}\)

6. Calculate \(\mathbf{PA} \cdot \mathbf{d} = (-1)(1) + (-4)(2) + (-4)(3) = -1 - 8 - 12 = -21\).

7. Calculate \(\mathbf{d} \cdot \mathbf{d} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\).

8. The projection is:

\(\text{proj}_{\mathbf{d}} \mathbf{PA} = \frac{-21}{14} \mathbf{d} = -\frac{3}{2} \mathbf{d}\)

9. The perpendicular vector \(\mathbf{v}\) from \(P\) to \(l\) is:

\(\mathbf{v} = \mathbf{PA} - \text{proj}_{\mathbf{d}} \mathbf{PA}\)

10. Calculate \(\mathbf{v} = (-\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}) - \left(-\frac{3}{2}(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\right)\).

11. Simplify \(\mathbf{v} = -\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} + \frac{3}{2}\mathbf{i} + 3\mathbf{j} + \frac{9}{2}\mathbf{k}\).

12. \(\mathbf{v} = \left(-1 + \frac{3}{2}\right)\mathbf{i} + (-4 + 3)\mathbf{j} + \left(-4 + \frac{9}{2}\right)\mathbf{k}\).

13. \(\mathbf{v} = \frac{1}{2}\mathbf{i} - \mathbf{j} + \frac{1}{2}\mathbf{k}\).

14. The length of \(\mathbf{v}\) is:

\(\|\mathbf{v}\| = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + 1 + \frac{1}{4}} = \sqrt{\frac{3}{2}}\)

15. \(\|\mathbf{v}\| \approx 1.22\) to 3 significant figures.

(i) To show that the lines do not intersect, equate the components of the general points on l and m:

\(For l: r = 3i − j − 2k + λ(−i + j + 4k)\)

\(For m: r = 4i + 4j − 3k + μ(2i + j − 2k)\)

Equate at least two pairs of components and solve for λ or μ:

\(1. 3 − λ = 4 + 2μ\)

\(2. −1 + λ = 4 + μ\)

Solving these gives λ = \(\frac{3}{2}\) or μ = −\(\frac{7}{2}\).

Verify that not all three pairs of equations are satisfied, confirming the lines do not intersect.

(ii) To calculate the acute angle between the directions of the lines, find the direction vectors:

\(For l: d₁ = −i + j + 4k\)

\(For m: d₂ = 2i + j − 2k\)

\(Calculate the scalar product: d₁ ⋅ d₂ = (−1)(2) + (1)(1) + (4)(−2) = −2 + 1 − 8 = −9\)

Calculate the moduli: |d₁| = \(\sqrt{(-1)^2 + 1^2 + 4^2} = \sqrt{18}\)

|d₂| = \(\sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\)

Divide the scalar product by the product of the moduli and evaluate the inverse cosine:

\(\cos \theta = \frac{-9}{\sqrt{18} \times 3} = \frac{-9}{9\sqrt{2}} = -\frac{1}{\sqrt{2}}\)

\(\theta = \cos^{-1}(-\frac{1}{\sqrt{2}}) = 45^{\circ}\) or \(\frac{1}{4} \pi\) (0.785 radians).

To find the vector equation of the line passing through points A and B, we use the position vectors:

\(\overrightarrow{OA} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\)

\(\overrightarrow{OB} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}\)

The direction vector \(\overrightarrow{AB}\) is given by:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + \mathbf{j} + \mathbf{k}) - (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})\)

\(= 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\)

The vector equation of the line through A and B is:

\(\mathbf{r} = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) + \lambda(2\mathbf{i} + 3\mathbf{j} - \mathbf{k})\)

Equating the components of the line \(l\) and the line through A and B:

\(2 + \mu = 1 + 2\lambda\)

\(1 - 2\mu = -2 + 3\lambda\)

\(m - 4\mu = 2 - \lambda\)

Solving these equations, we find \(\lambda = \frac{5}{7}\) and \(\mu = \frac{3}{7}\).

Substituting \(\mu = \frac{3}{7}\) into the third equation:

\(m - 4\left(\frac{3}{7}\right) = 2 - \frac{5}{7}\)

\(m - \frac{12}{7} = \frac{9}{7}\)

\(m = \frac{9}{7} + \frac{12}{7} = 3\)

Thus, the value of \(m\) is \(3\).

(a) To find the values of \(a\), \(b\), and \(c\), we use the fact that lines l and m are perpendicular. The direction vectors of l and m are \((c, -2, 4)\) and \((2, -3, 1)\) respectively. The scalar product of these vectors must be zero:

\(2c + 6 + 4 = 0\)

Solving gives \(c = -5\).

Since P lies on l, substitute \(\mathbf{r} = 4\mathbf{i} + 7\mathbf{j} - 2\mathbf{k}\) into the equation of l:

\(4 = a + \lambda c\)

\(7 = 3 - 2\lambda\)

\(-2 = b + 4\lambda\)

Solving these equations gives \(a = -6\), \(b = 6\), and \(\lambda = 2\).

(b) To find the position vector of R, first find the position vector of Q by setting the perpendicular from P to m:

\(\mathbf{OQ} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using the ratio \(PQ : QR = 2 : 3\), find \(\mathbf{OR}\):

\(\mathbf{OR} = \mathbf{OP} + \frac{5}{2} \mathbf{PQ}\)

\(\mathbf{OR} = \left(4, 7, -2\right) + \frac{5}{2} \left(-5, -2, 4\right)\)

\(\mathbf{OR} = \left(\frac{17}{2}, \frac{1}{2}, 8\right)\)

To find the foot of the perpendicular from A to the line l, consider a general point P on l with parameter \(\lambda\), given by:

\(\overrightarrow{OP} = (9 + 3\lambda)\mathbf{i} + (-1 - \lambda)\mathbf{j} + (8 + 2\lambda)\mathbf{k}\).

The vector \(\overrightarrow{AP}\) is:

\(\overrightarrow{AP} = (8 + 3\lambda)\mathbf{i} + (-3 - \lambda)\mathbf{j} + (4 + 2\lambda)\mathbf{k}\).

For \(\overrightarrow{AP}\) to be perpendicular to the direction vector of l, \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\), their dot product must be zero:

\((8 + 3\lambda) \cdot 3 + (-3 - \lambda) \cdot (-1) + (4 + 2\lambda) \cdot 2 = 0\).

Simplifying, we get:

\(24 + 9\lambda + 3 + \lambda + 8 + 4\lambda = 0\).

\(14\lambda + 35 = 0\).

\(\lambda = -\frac{5}{2}\).

Substitute \(\lambda = -\frac{5}{2}\) back into \(\overrightarrow{OP}\) to find the foot of the perpendicular:

\(\overrightarrow{OP} = \left(9 + 3\left(-\frac{5}{2}\right)\right)\mathbf{i} + \left(-1 - \left(-\frac{5}{2}\right)\right)\mathbf{j} + \left(8 + 2\left(-\frac{5}{2}\right)\right)\mathbf{k}\).

\(\overrightarrow{OP} = \frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 3\mathbf{k}\).

The reflection of A in l is found by using the formula:

\(\overrightarrow{OR} = 2\overrightarrow{OP} - \overrightarrow{OA}\).

\(\overrightarrow{OR} = 2\left(\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} + 3\mathbf{k}\right) - (\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).

\(\overrightarrow{OR} = (3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).

\(\overrightarrow{OR} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).

Express the general point on the line l in component form:

\((1 + 2\lambda, 2 - \lambda, 1 + \lambda)\).

The distance from the origin is given by:

\(\sqrt{(1 + 2\lambda)^2 + (2 - \lambda)^2 + (1 + \lambda)^2} = \sqrt{10}\).

Squaring both sides, we have:

\((1 + 2\lambda)^2 + (2 - \lambda)^2 + (1 + \lambda)^2 = 10\).

Expanding and simplifying:

\(1 + 4\lambda + 4\lambda^2 + 4 - 4\lambda + \lambda^2 + 1 + 2\lambda + \lambda^2 = 10\).

Combine like terms:

\(6\lambda^2 + 2\lambda + 6 = 10\).

Rearrange to form a quadratic equation:

\(6\lambda^2 + 2\lambda - 4 = 0\).

Divide the entire equation by 2:

\(3\lambda^2 + \lambda - 2 = 0\).

Using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), \(c = -2\):

\(\lambda = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm 5}{6}\).

This gives \(\lambda = \frac{2}{3}\) or \(\lambda = -1\).

Substitute back to find the position vectors:

For \(\lambda = -1\):

\((1 + 2(-1), 2 - (-1), 1 + (-1)) = (-1, 3, 0)\).

For \(\lambda = \frac{2}{3}\):

\((1 + 2\left(\frac{2}{3}\right), 2 - \left(\frac{2}{3}\right), 1 + \left(\frac{2}{3}\right)) = \left(\frac{7}{3}, \frac{4}{3}, \frac{5}{3}\right)\).

(i) The vector equation of the line through A and B is \(\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} - \mathbf{j} + 2\mathbf{k})\).

Equating components with the line \(l\):

\(1 + \lambda = 2 - \mu\)

\(1 - \lambda = -2 + 2\mu\)

\(1 + 2\lambda = -1 + \mu\)

Solving these equations, we find \(\lambda = -1\) and \(\mu = 2\), but the third equation is not satisfied, showing no intersection.

(ii) To find the perpendicular distance from A to \(l\), consider a point \(P\) on \(l\) given by \((1-\mu)\mathbf{i} + (-3 + 2\mu)\mathbf{j} + (-2 + \mu)\mathbf{k}\).

Calculate \(\overrightarrow{AP}\) and find \(\mu\) such that \(\overrightarrow{AP}\) is perpendicular to the direction vector of \(l\), \(-\mathbf{i} + 2\mathbf{j} + \mathbf{k}\).

Solving gives \(\mu = \frac{3}{2}\).

Substitute \(\mu = \frac{3}{2}\) into \(\overrightarrow{AP}\) and find its magnitude: \(\frac{1}{\sqrt{2}}\).

To find the position vector of D, we use the property of a parallelogram that opposite sides are equal and parallel. Therefore, \(\overrightarrow{AB} = \overrightarrow{CD}\).

First, calculate \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (0\mathbf{i} + 4\mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = -\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Since \(\overrightarrow{CD} = \overrightarrow{AB}\), we have:

\(\overrightarrow{CD} = -\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Now, find \(\overrightarrow{OD}\):

\(\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{CD} = (2\mathbf{i} + 5\mathbf{j} - \mathbf{k}) + (-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = 3\mathbf{i} + 3\mathbf{j} + \mathbf{k}\).

To verify that ABCD is a rhombus, we need to show that all sides are equal in length. Calculate the magnitudes of \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

\(|\overrightarrow{AB}| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\).

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (2\mathbf{i} + 5\mathbf{j} - \mathbf{k}) - (0\mathbf{i} + 4\mathbf{j} + \mathbf{k}) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\).

\(|\overrightarrow{BC}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\).

Since \(|\overrightarrow{AB}| = |\overrightarrow{BC}|\), ABCD has a pair of adjacent sides that are equal, confirming it is a rhombus.

First, find the vector equation of the line passing through A and B. The direction vector \(\overrightarrow{AB}\) is given by:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\mathbf{i} + \mathbf{j} + 5\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\)

The vector equation of the line through A and B is:

\(\mathbf{r} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k} + \lambda(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\)

Equate the components of the general points on both lines:

\(2 - \lambda = 1 + 3\mu\)

\(-1 + 2\lambda = 1 + \mu\)

\(3 + 2\lambda = 2 - \mu\)

Solve these equations for \(\lambda\) and \(\mu\):

From the first equation: \(\lambda = 1 - 3\mu\)

Substitute into the second equation:

\(-1 + 2(1 - 3\mu) = 1 + \mu\)

\(-1 + 2 - 6\mu = 1 + \mu\)

\(-6\mu - \mu = 1\)

\(-7\mu = 0\)

\(\mu = 0\)

Substitute \(\mu = 0\) into \(\lambda = 1 - 3\mu\):

\(\lambda = 1\)

Check the third equation:

\(3 + 2(1) = 2 - 0\)

\(5 \neq 2\)

Since not all three equations are satisfied, the lines do not intersect.

(i) The direction vector of \(l_1\) is obtained from the points \((0, 1, 5)\) and \((2, -2, 1)\), which is \(\begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}\).

The direction vector of \(l_2\) is \(\begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}\).

Since the direction vectors are not parallel, the lines are not parallel.

Express the general point of \(l_1\) as \((2\lambda, 1 - 3\lambda, 5 - 4\lambda)\) and \(l_2\) as \((7 + \mu, 1 + 2\mu, 1 + 5\mu)\).

Equate at least two pairs of components and solve for \(\lambda\) and \(\mu\). Verify that all three component equations are not satisfied, confirming the lines are skew.

(ii) The direction vector of \(l_2\) is \(\begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}\) and the direction of the \(x\)-axis is \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).

The scalar product is \(1 \times 1 + 2 \times 0 + 5 \times 0 = 1\).

The magnitudes are \(\sqrt{1^2 + 2^2 + 5^2} = \sqrt{30}\) and \(\sqrt{1^2} = 1\).

The cosine of the angle is \(\frac{1}{\sqrt{30}}\).

The angle is \(\cos^{-1}\left(\frac{1}{\sqrt{30}}\right) \approx 79.5^\circ\) or \(1.39\) radians.

(i) To show that the lines intersect, equate the vector equations:

\(1 + \lambda = a + \mu\)

\(4 = 2 + 2\mu\)

\(-2 + 3\lambda = -2 + 3a\mu\)

From the second equation, solve for \(\mu\):

\(2 = 2\mu \Rightarrow \mu = 1\)

Substitute \(\mu = 1\) into the first equation:

\(1 + \lambda = a + 1 \Rightarrow \lambda = a\)

Check the third equation:

\(-2 + 3a = -2 + 3a\)

All equations are satisfied, so the lines intersect for all \(a\).

(ii) The point of intersection is \((a + 1, 4, 3a - 2)\).

The distance from the origin is given by:

\(\sqrt{(a+1)^2 + 4^2 + (3a-2)^2} = 9\)

\((a+1)^2 + 16 + (3a-2)^2 = 81\)

\(a^2 + 2a + 1 + 16 + 9a^2 - 12a + 4 = 81\)

\(10a^2 - 10a + 21 = 81\)

\(10a^2 - 10a - 60 = 0\)

\(a^2 - a - 6 = 0\)

Factor the quadratic:

\((a - 3)(a + 2) = 0\)

Thus, \(a = 3\) or \(a = -2\).

To find the length of the perpendicular from point A to the line l, we first need to find the vector AP where P is a point on the line l.

The line l is given by r = 4i - 9j + 9k + \(\lambda (-2i + j - 2k)\).

Let P be a point on the line, then the position vector of P is r = (4 - 2\(\lambda\))i + (-9 + \(\lambda\))j + (9 - 2\(\lambda\))k.

The position vector of A is 3i + 8j + 5k.

Thus, AP = ((4 - 2\(\lambda\)) - 3)i + ((-9 + \(\lambda\)) - 8)j + ((9 - 2\(\lambda\)) - 5)k.

Simplifying, AP = (1 - 2\(\lambda\))i + (-17 + \(\lambda\))j + (4 - 2\(\lambda\))k.

To find the perpendicular distance, we need to minimize the magnitude of AP.

The magnitude of AP is \(\sqrt{(1 - 2\lambda)^2 + (-17 + \lambda)^2 + (4 - 2\lambda)^2}\).

Expanding and simplifying, we get:

\((1 - 2\lambda)^2 = 1 - 4\lambda + 4\lambda^2\)

\((-17 + \lambda)^2 = 289 - 34\lambda + \lambda^2\)

\((4 - 2\lambda)^2 = 16 - 16\lambda + 4\lambda^2\)

Adding these, we get:

\(5\lambda^2 - 54\lambda + 306\)

To minimize, set the derivative with respect to \(\lambda\) to zero:

\(10\lambda - 54 = 0\)

\(\lambda = 5.4\)

Substitute \(\lambda = 5.4\) back into the expression for the magnitude:

\(\sqrt{5(5.4)^2 - 54(5.4) + 306} = 15\)

Thus, the length of the perpendicular from A to l is 15.

(i) To find the cosine of angle BAC, we first find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\mathbf{i} + 4\mathbf{j} + \mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\)

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \cdot (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) = 2 + 6 + 12 = 20\)

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\) and \(|\overrightarrow{AC}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49} = 7\)

Thus, \(\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{20}{3 \times 7} = \frac{20}{21}\)

(ii) To find the area of triangle ABC, we use the sine of angle BAC:

\(\sin \angle BAC = \sqrt{1 - \left(\frac{20}{21}\right)^2} = \sqrt{\frac{41}{441}} = \frac{\sqrt{41}}{21}\)

The area is \(\frac{1}{2} |\overrightarrow{AB}| |\overrightarrow{AC}| \sin \angle BAC = \frac{1}{2} \times 3 \times 7 \times \frac{\sqrt{41}}{21} = \frac{1}{2} \sqrt{41}\)

To find the intersection, equate the parametric equations of the lines:

\(\begin{pmatrix} 5 + s \\ 1 - s \\ -4 + 3s \end{pmatrix} = \begin{pmatrix} p + 2t \\ 4 + 5t \\ -2 - 4t \end{pmatrix}\).

Equating components, we get:

1. \(5 + s = p + 2t\)

2. \(1 - s = 4 + 5t\)

3. \(-4 + 3s = -2 - 4t\)

From equation 2: \(s = -3 - 5t\).

Substitute \(s = -3 - 5t\) into equation 3:

\(-4 + 3(-3 - 5t) = -2 - 4t\)

\(-4 - 9 - 15t = -2 - 4t\)

\(-13 - 15t = -2 - 4t\)

\(-11 = 11t\)

\(t = -1\).

Substitute \(t = -1\) into \(s = -3 - 5t\):

\(s = -3 - 5(-1) = 2\).

Substitute \(s = 2\) and \(t = -1\) into equation 1:

\(5 + 2 = p + 2(-1)\)

\(7 = p - 2\)

\(p = 9\).

Thus, the value of \(p\) is 9. The intersection point is:

\(\begin{pmatrix} 5 + 2 \\ 1 - 2 \\ -4 + 3(2) \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 2 \end{pmatrix}\).

(a) To find the vector equation for the line passing through A and B, use the position vectors of A and B:

\(\mathbf{r} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 3\mathbf{j} + 3\mathbf{k})\)

Equate the components of the general points on their line and \(l\):

\(\begin{align*} 1 + \lambda &= 1 + 2\mu, \\ 2 - 3\lambda &= -1 - 3\mu, \\ -2 + 3\lambda &= 3 + 4\mu. \end{align*}\)

Solving these equations gives \(\lambda = -1\) and \(\mu = -2\). However, substituting these values into the third equation does not satisfy it, confirming the lines do not intersect.

(b) For a general point \(P\) on \(l\), \(\mathbf{P} = -3\mathbf{j} + 5\mathbf{k} + \mu(2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k})\).

Calculate the scalar product of \(\overrightarrow{AP}\) and a direction vector for \(l\) and equate to zero:

\(4\mu + (9 + 9\mu) + (20 + 16\mu) = 0\)

Solving gives \(\mu = -1\).

Substitute \(\mu = -1\) back into the equation for \(P\) to find the position vector:

\(\mathbf{P} = -\mathbf{i} + 2\mathbf{j} - \mathbf{k}\)

(i) Express the general point of l or m in component form: l: (3 - λ, -2 + 2λ, 1 + λ) and m: (4 + aμ, 4 + bμ, 2 - μ).

\(Equate components and eliminate either λ or μ from a pair of equations. For example, equate the i components: 3 - λ = 4 + aμ.\)

Eliminate the other parameter and obtain an equation in a and b. For example, equate the k components: 1 + λ = 2 - μ.

\(From these, solve to obtain 2a - b = 4.\)

\((ii) Using the correct process, equate the scalar product of the direction vectors to zero: (-1, 2, 1) ⋅ (a, b, -1) = 0.\)

\(This gives -a + 2b - 1 = 0.\)

\(Solve the simultaneous equations for a and b: 2a - b = 4 and -a + 2b - 1 = 0.\)

\(Obtain a = 3, b = 2.\)

(iii) Substitute found values in component equations and solve for λ or μ.

\(Obtain the position vector i + 2j + 3k from either λ = 2 or μ = -1.\)

To find the perpendicular distance from point P to the line l, we can use the vector approach.

First, find the vector PA where A is a point on the line. Let A be the point corresponding to \(\lambda = 0\), so \(A = \begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix}\).

Then, \(\mathbf{PA} = \begin{pmatrix} 1 - (-1) \\ 3 - 4 \\ -4 - 11 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -15 \end{pmatrix}\).

The direction vector of the line is \(\mathbf{d} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\).

Use the cross product to find the area of the parallelogram formed by PA and d:

\(\mathbf{PA} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -15 \\ 2 & 1 & 3 \end{vmatrix} = \begin{pmatrix} (-1)(3) - (-15)(1) \\ -((2)(3) - (-15)(2)) \\ (2)(1) - (-1)(2) \end{pmatrix} = \begin{pmatrix} 12 \\ -36 \\ 4 \end{pmatrix}\).

The magnitude of this cross product is \(\sqrt{12^2 + (-36)^2 + 4^2} = \sqrt{144 + 1296 + 16} = \sqrt{1456}\).

The magnitude of the direction vector \(\mathbf{d}\) is \(\sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).

The perpendicular distance is given by \(\frac{\sqrt{1456}}{\sqrt{14}} = \sqrt{104} \approx 10.2\).

(i) The vector \(\overrightarrow{AB}\) is given by:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\)

Since \(\overrightarrow{AP} = \lambda \overrightarrow{AB}\), we have:

\(\overrightarrow{AP} = \lambda (2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = 2\lambda \mathbf{i} + 2\lambda \mathbf{j} - 2\lambda \mathbf{k}\)

Thus, \(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) + (2\lambda \mathbf{i} + 2\lambda \mathbf{j} - 2\lambda \mathbf{k})\).

Therefore, \(\overrightarrow{OP} = (1 + 2\lambda)\mathbf{i} + (2 + 2\lambda)\mathbf{j} + (2 - 2\lambda)\mathbf{k}\).

(ii) To find \(\lambda\) such that \(OP\) bisects \(\angle AOB\), equate the expressions for \(\cos AOP\) and \(\cos BOP\):

\(\frac{9 + 2\lambda}{3\sqrt{9 + 4\lambda + 12\lambda^2}} = \frac{11 + 14\lambda}{5\sqrt{9 + 4\lambda + 12\lambda^2}}\)

Solving this equation gives \(\lambda = \frac{3}{8}\).

(iii) Verify \(AP : PB = OA : OB\):

\(AP = \lambda \times AB = \frac{3}{8} \times \sqrt{12}\)

\(PB = (1 - \lambda) \times AB = \frac{5}{8} \times \sqrt{12}\)

\(OA = \sqrt{9}\), \(OB = \sqrt{25}\)

\(\frac{AP}{PB} = \frac{3}{5} = \frac{OA}{OB}\)

Thus, the verification holds.

(i) To prove that the lines l and m do not intersect, express the general point of l in component form: \((2 + \lambda, -\lambda, 1 + 2\lambda)\) and the general point of m as \((\mu, 2 + 2\mu, 6 - 2\mu)\). Equate the components and solve for \(\lambda\) and \(\mu\). Possible solutions for \(\lambda\) are -2, \(\frac{1}{4}\), 7, and for \(\mu\) are 0, \(\frac{1}{4}\), -\(\frac{1}{2}\). Verify that all three component equations are not satisfied simultaneously, confirming that the lines do not intersect.

Alternatively, use the scalar triple product: \((2i - 2j - 5k) \cdot ((i - j + 2k) \times (i + 2j - 2k))\). Calculate the determinant to obtain a non-zero value, e.g., -27, indicating the lines do not intersect.

(ii) To find the acute angle between the directions of l and m, calculate the scalar product of the direction vectors \((i - j + 2k)\) and \((i + 2j - 2k)\). The scalar product is -3. Calculate the moduli of the direction vectors: \(\sqrt{6}\) and \(\sqrt{9}\). Use the formula \(\cos \theta = \frac{-3}{\sqrt{6} \times \sqrt{9}}\) to find \(\theta\). The acute angle is approximately 47.1° or 0.822 radians.

(i) The direction vector of line AB is \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).

Thus, the vector equation of the line AB is \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\).

(ii) Since OP is perpendicular to AB, the dot product \(\overrightarrow{OP} \cdot \overrightarrow{AB} = 0\).

Let \(\overrightarrow{OP} = (1 + 2\lambda)\mathbf{i} + (2 + 2\lambda)\mathbf{j} + (2 - 2\lambda)\mathbf{k}\).

Then, \(\overrightarrow{OP} \cdot \overrightarrow{AB} = (1 + 2\lambda)(2) + (2 + 2\lambda)(2) + (2 - 2\lambda)(-2) = 0\).

Simplifying, \(2 + 4\lambda + 4 + 4\lambda - 4 + 4\lambda = 0\).

Thus, \(12\lambda + 2 = 0\), giving \(\lambda = -\frac{1}{6}\).

Substitute \(\lambda = -\frac{1}{6}\) into \(\overrightarrow{OP}\):

\(\overrightarrow{OP} = (1 + 2(-\frac{1}{6}))\mathbf{i} + (2 + 2(-\frac{1}{6}))\mathbf{j} + (2 - 2(-\frac{1}{6}))\mathbf{k}\).

\(\overrightarrow{OP} = \frac{2}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} + \frac{7}{3}\mathbf{k}\).

(i) Express the general point of l in component form: \((1 + s, 1 - s, 1 + 2s)\).

Express the general point of m in component form: \((4 + 2t, 6 + 2t, 1 + t)\).

Equate the components:

\(1 + s = 4 + 2t\)

\(1 - s = 6 + 2t\)

\(1 + 2s = 1 + t\)

Solving these equations, we find \(s = -1\) or \(t = -2\).

Verify that all three component equations are satisfied with these values.

(ii) The direction vector of l is \((1, -1, 2)\) and of m is \((2, 2, 1)\).

Calculate the scalar product: \(1 \times 2 + (-1) \times 2 + 2 \times 1 = 2 - 2 + 2 = 2\).

Calculate the moduli: \(\sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}\) and \(\sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3\).

Divide the scalar product by the product of the moduli: \(\frac{2}{\sqrt{6} \times 3}\).

Evaluate the inverse cosine: \(\cos^{-1}\left(\frac{2}{3\sqrt{6}}\right)\).

The acute angle between the lines is \(74.2^\circ\) (or 1.30 radians).

(i) The position vector of the midpoint \(M\) of \(AB\) is:

\(\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2}((\mathbf{i} - \mathbf{k}) + (3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k})) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\)

The point \(N\) divides \(AC\) in the ratio \(2:1\), so:

\(\overrightarrow{ON} = \frac{2\overrightarrow{OC} + \overrightarrow{OA}}{3} = \frac{2(4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) + (\mathbf{i} - \mathbf{k})}{3} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}\)

The direction vector of \(MN\) is:

\(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) - (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = \mathbf{i} - 3\mathbf{j} + 3\mathbf{k}\)

Thus, the vector equation of the line \(MN\) is:

\(\mathbf{r} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 3\mathbf{j} + 3\mathbf{k})\)

(ii) The equation of line \(BC\) is:

\(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} + \mu(\mathbf{i} - 5\mathbf{j} + 5\mathbf{k})\)

Setting the equations of \(MN\) and \(BC\) equal gives:

\(2\mathbf{i} + \mathbf{j} - 2\mathbf{k} + \lambda(\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}) = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} + \mu(\mathbf{i} - 5\mathbf{j} + 5\mathbf{k})\)

Equating components, we solve for \(\lambda\) and \(\mu\):

\(\lambda = 3, \quad \mu = 2\)

Substituting \(\mu = 2\) into the equation of \(BC\):

\(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} + 2(\mathbf{i} - 5\mathbf{j} + 5\mathbf{k}) = 5\mathbf{i} - 8\mathbf{j} + 7\mathbf{k}\)

Thus, the position vector of \(P\) is \(5\mathbf{i} - 8\mathbf{j} + 7\mathbf{k}\).

(i) The vector equation for the line through A and B is \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j})\). Equating components with the line \(l\), we have:

\(1 + s = 1 - 2t\)

\(2 - s = 5 + t\)

\(3 = 2 - t\)

Solving these gives \(s = -6, t = 3\) for the first equation, \(s = 2, t = -1\) for the second, and \(t = -1\) for the third. The values do not satisfy all equations simultaneously, so \(l\) does not intersect the line through A and B.

(ii) The direction vector for \(AP\) is \((-2t, 3 + t, -1 - t)\). The cosine of angle \(PAB\) is given by:

\(\cos 60^{\circ} = \frac{\overrightarrow{AP} \cdot \overrightarrow{AB}}{|\overrightarrow{AP}||\overrightarrow{AB}|}\)

Expanding the scalar product and expressing the product of the moduli in terms of \(t\), we obtain the equation \(3t^2 + 7t + 2 = 0\).

Solving the quadratic equation, we find roots \(t = -2\) and \(t = -\frac{1}{3}\). The position vector of P for \(t = -2\) is \(5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\), which is valid as it satisfies the angle condition.

(i) The vector equation of the line l passing through A and parallel to OB is given by:

\(\mathbf{r} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\)

where \(\lambda\) is a scalar.

(ii) To find the position vector of N, express \(\overrightarrow{BN}\) in terms of \(\lambda\):

\(\overrightarrow{BN} = \begin{pmatrix} -1 + 3\lambda \\ 3 - \lambda \\ 5 - 4\lambda \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\)

Equate its scalar product with \(\begin{pmatrix} 3 \\ -1 \\ -4 \end{pmatrix}\) to zero and solve for \(\lambda\):

\((3)(-1 + 3\lambda) + (-1)(3 - \lambda) + (-4)(5 - 4\lambda) = 0\)

Simplifying gives \(\lambda = 2\).

Substitute \(\lambda = 2\) back to find \(\overrightarrow{ON}\):

\(\overrightarrow{ON} = \begin{pmatrix} -1 + 6 \\ 3 - 2 \\ 5 - 8 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \\ -3 \end{pmatrix}\)

To show \(BN = 3\), calculate the magnitude of \(\overrightarrow{BN}\):

\(\overrightarrow{BN} = \begin{pmatrix} 5 - 3 \\ 1 + 1 \\ -3 + 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}\)

\(BN = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\)

First, find the direction vector for the line through A and B:

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

The equation of the line through A and B is:

\(\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).

Equate the vector equations of the lines to find if they intersect:

\(4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} + s(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).

Equating components, we get:

1. \(4 + s = 2 - t\)

2. \(-2 + 2s = 2 + 2t\)

3. \(2 + s = 1 + 2t\)

Solving these equations, we find:

From equation 1: \(s + t = -2\)

From equation 2: \(s - t = 2\)

Adding these, \(2s = 0 \Rightarrow s = 0\).

Substituting \(s = 0\) into \(s + t = -2\), we get \(t = -2\).

Substitute \(s = 0\) and \(t = -2\) into equation 3:

\(2 + 0 \neq 1 + 2(-2) \Rightarrow 2 \neq -3\).

Since the equations are inconsistent, the lines do not intersect.

(a) To find the obtuse angle between \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\), calculate the scalar product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3)(1) + (-1)(2) + (2)(-3) = 3 - 2 - 6 = -5\)

Calculate the magnitudes:

\(|\overrightarrow{OA}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{14}\)

\(|\overrightarrow{OB}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}\)

Use the cosine formula:

\(\cos \theta = \frac{-5}{\sqrt{14} \times \sqrt{14}} = \frac{-5}{14}\)

\(\theta = \cos^{-1}\left(\frac{-5}{14}\right) \approx 110.9^\circ \text{ or } 1.94 \text{ radians}\)

(b) The vector equation for the line \(l\) through points \(A\) and \(B\) is:

\(\mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \mu_1 \begin{pmatrix} 1 - 3 \\ 2 + 1 \\ -3 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \mu_1 \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}\)

(c) For the line through \(C\) and \(D\):

\(\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 5 - 1 \\ -6 + 2 \\ 11 - 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix}\)

Equate components and solve for \(\lambda\) and \(\mu_1\):

\(3 + 2\mu_1 = 1 + 4\lambda\)

\(-1 - 3\mu_1 = -2 - 4\lambda\)

\(2 + 5\mu_1 = 5 + 6\lambda\)

Solve to find \(\mu_1 = 3\) and \(\lambda = 2\).

Substitute back to find the position vector:

\(\mathbf{r} = \begin{pmatrix} 9 \\ -10 \\ 17 \end{pmatrix}\)

(i) Express the general point of l or m in component form:

For l: \((2 + s, -1 + s, 4 - s)\)

For m: \((-2 - 2t, 2 + t, 1 + t)\)

Equate at least two pairs of components and solve for \(s\) or \(t\).

Possible solutions: \(s = \frac{2}{3}, 10, 3\) or \(t = \frac{7}{3}, -7, 0\).

Verify that all three component equations are not satisfied, hence lines do not intersect.

(ii) Express \(\overrightarrow{PQ}\) in terms of \(s\):

\(\overrightarrow{PQ} = -s\mathbf{i} + (1 - s)\mathbf{j} + (-5 + s)\mathbf{k}\)

Equate its scalar product with a direction vector for l to zero:

\((-s)(1) + (1-s)(1) + (-5+s)(-1) = 0\)

Solve for \(s\): \(s = 2\)

\(\mathbf{OP} = 4\mathbf{i} + \mathbf{j} + 2\mathbf{k}\)

(iii) Show that Q is on m with parameter \(t = -2\):

Substitute \(t = -2\) into m's equation:

\(\mathbf{r} = -2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + (-2)(-2\mathbf{i} + \mathbf{j} + \mathbf{k}) = 2\mathbf{i} - \mathbf{k}\)

Verify \(PQ\) is perpendicular to m:

\((-2\mathbf{i} - \mathbf{j} - 3\mathbf{k}) \cdot (-2\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0\)

Express the general points on lines l and m in component form:

Line l: \((1 + 2s, s, -2 + 3s)\)

Line m: \((6 + t, -5 - 2t, 4 + t)\)

Equate the components to find \(s\) and \(t\):

1. \(1 + 2s = 6 + t\)

2. \(s = -5 - 2t\)

3. \(-2 + 3s = 4 + t\)

From equation (1):

\(t = 2s - 5\)

Substitute \(t = 2s - 5\) into equation (2):

\(s = -5 - 2(2s - 5)\)

\(s = -5 - 4s + 10\)

\(5s = 5\)

\(s = 1\)

Substitute \(s = 1\) into \(t = 2s - 5\):

\(t = 2(1) - 5 = -3\)

Verify with equation (3):

\(-2 + 3(1) = 4 + (-3)\)

\(1 = 1\)

All equations are satisfied. The position vector of the intersection point is:

\(3\mathbf{i} + \mathbf{j} + \mathbf{k}\)

(i) Find direction vectors for \(\overrightarrow{AB}\) and \(\overrightarrow{CD}\):

\(\overrightarrow{AB} = (5 - 4)\mathbf{i} + (-2 - 0)\mathbf{j} + (-2 - 1)\mathbf{k} = \mathbf{i} - 2\mathbf{j} - 3\mathbf{k}\)

\(\overrightarrow{CD} = (-1 - 1)\mathbf{i} + (0 - 1)\mathbf{j} + (-4 - 0)\mathbf{k} = -2\mathbf{i} - \mathbf{j} - 4\mathbf{k}\)

Calculate the cosine of the angle:

\(\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|}\)

\(\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(-2) + (-2)(-1) + (-3)(-4) = -2 + 2 + 12 = 12\)

\(|\overrightarrow{AB}| = \sqrt{1^2 + (-2)^2 + (-3)^2} = \sqrt{14}\)

\(|\overrightarrow{CD}| = \sqrt{(-2)^2 + (-1)^2 + (-4)^2} = \sqrt{21}\)

\(\cos \theta = \frac{12}{\sqrt{14} \times \sqrt{21}} = \frac{12}{\sqrt{294}}\)

\(\theta = \cos^{-1}\left(\frac{12}{\sqrt{294}}\right) \approx 45.6^{\circ} \text{ or } 0.796 \text{ radians}\)

(ii) Use line equations:

\(\overrightarrow{r} = \overrightarrow{OA} + \lambda \overrightarrow{AB} = (4 + \lambda)\mathbf{i} + (-2\lambda)\mathbf{j} + (1 - 3\lambda)\mathbf{k}\)

\(\overrightarrow{r} = \overrightarrow{OC} + \mu \overrightarrow{CD} = (1 - 2\mu)\mathbf{i} + (1 - \mu)\mathbf{j} + (-4\mu)\mathbf{k}\)

Equate components and solve:

\(4 + \lambda = 1 - 2\mu\)

\(-2\lambda = 1 - \mu\)

\(1 - 3\lambda = -4\mu\)

Solving gives \(\lambda = -1\), \(\mu = 1\), showing lines intersect.

(iii) Find \(\overrightarrow{PQ}\) for a general point \(Q\) on \(AB\):

\(\overrightarrow{PQ} = (3 - 5\lambda)\mathbf{i} + (5 + 2\lambda)\mathbf{j} + (6 + 3\lambda)\mathbf{k}\)

Calculate \(\overrightarrow{PQ} \cdot \overrightarrow{AB}\) and equate to zero:

\((3 - 5\lambda) + (-4\lambda) + (-18 - 9\lambda) = 0\)

Solve for \(\lambda\):

\(-18\lambda = -18 \Rightarrow \lambda = -2\)

\(\overrightarrow{PQ} = -5\mathbf{i} + 4\mathbf{j} + \mathbf{k}\)

\(|\overrightarrow{PQ}| = \sqrt{(-5)^2 + 4^2 + 1^2} = \sqrt{42}\)

\(\text{Distance} = \frac{|\overrightarrow{PQ} \times \overrightarrow{AB}|}{|\overrightarrow{AB}|} = \sqrt{3}\)

(a) To find the position vector of M, the midpoint of AC, use the midpoint formula:

\(\overrightarrow{OM} = \frac{1}{2} (\overrightarrow{OA} + \overrightarrow{OC}) = \frac{1}{2} \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + \begin{pmatrix} 4 \\ -3 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}.\)

\(For N, given BN = 2NC, we find:\)

\(\overrightarrow{ON} = \overrightarrow{OB} + \frac{2}{3}(\overrightarrow{OC} - \overrightarrow{OB}) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \frac{2}{3} \begin{pmatrix} 3 \\ -3 \\ -3 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -1 \end{pmatrix}.\)

(b) The vector equation for the line through M and N is:

\(\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix}.\)

(c) The vector equation for the line through A and B is:

\(\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -5 \\ -1 \end{pmatrix}.\)

Equating components of AB and MN gives:

\(\lambda = -3\) or \(\mu = 2.\)

Substituting \(\mu = 2\) into the equation for AB gives:

\(\overrightarrow{OQ} = \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ -5 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 10 \\ 3 \end{pmatrix}.\)

(a) The direction vector of \(OA\) is \(\langle 1, 5, 6 \rangle\) and the direction vector of \(l\) is \(\langle -1, 2, 3 \rangle\). The scalar product is \(1(-1) + 5(2) + 6(3) = -1 + 10 + 18 = 27\).

The moduli are \(\sqrt{1^2 + 5^2 + 6^2} = \sqrt{62}\) and \(\sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{14}\).

The cosine of the angle is \(\cos \theta = \frac{27}{\sqrt{62} \cdot \sqrt{14}}\).

Thus, \(\theta = \cos^{-1} \left( \frac{27}{\sqrt{62} \cdot \sqrt{14}} \right) \approx 23.6^{\circ}\).

(b) Let \(P\) be the foot of the perpendicular from \(A\) to \(l\). The position vector of \(P\) is \(\mathbf{OP} = 4\mathbf{i} + \mathbf{k} + \lambda (-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\).

Set \(\mathbf{AP} \cdot \langle -1, 2, 3 \rangle = 0\) to find \(\lambda\). \(\mathbf{AP} = \langle 3 - \lambda, -5 + 2\lambda, -5 + 3\lambda \rangle\).

Solving \((3 - \lambda)(-1) + (-5 + 2\lambda)(2) + (-5 + 3\lambda)(3) = 0\) gives \(\lambda = 2\).

Thus, \(\mathbf{OP} = 2\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\).

(c) The reflection of \(A\) in \(l\) is \(\mathbf{A'} = \mathbf{OP} + (\mathbf{OP} - \mathbf{OA})\).

\(\mathbf{A'} = 2\mathbf{i} + 4\mathbf{j} + 7\mathbf{k} + (2\mathbf{i} + 4\mathbf{j} + 7\mathbf{k} - (\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}))\).

\(\mathbf{A'} = 3\mathbf{i} + 3\mathbf{j} + 8\mathbf{k}\).

(a) Express the general point of l or m in component form:

\((-1 + 2\lambda, 3 - \lambda, 4 - \lambda)\)

\((5 + a\mu, 4 + b\mu, 3 + \mu)\)

Equate components and eliminate either \(\lambda\) or \(\mu\):

\(\mu = \frac{2\lambda - 6}{a}\)

\(\mu = \frac{-1 - \lambda}{b}\)

Use \(\lambda = 1 - \mu\) to obtain \(2b - a = 4\).

(b) For perpendicular lines, the scalar product of direction vectors is zero:

\((2\mathbf{i} - \mathbf{j} - \mathbf{k}) \cdot (a\mathbf{i} + b\mathbf{j} + \mathbf{k}) = 0\)

\(2a - b - 1 = 0\)

Solve simultaneous equations:

\(2a - b = 1\)

\(2b - a = 4\)

Obtain \(a = 2, b = 3\).

(c) Substitute \(a = 2\) and \(b = 3\) into component equations and solve for \(\lambda\) or \(\mu\):

Obtain position vector \(3\mathbf{i} + \mathbf{j} + 2\mathbf{k}\).

(a) To find \(\overrightarrow{OM}\), note that M is the midpoint of DF. Since D is at \((0, 4, 2)\) and F is at \((2, 4, 4)\), M is at \(\left( \frac{0+2}{2}, \frac{4+4}{2}, \frac{2+4}{2} \right) = (1, 4, 3)\). Thus, \(\overrightarrow{OM} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

For \(\overrightarrow{MN}\), N divides AB in the ratio 3:1. A is at \((2, 0, 0)\) and B is at \((2, 4, 0)\), so N is at \(\left( 2, \frac{3 \times 0 + 1 \times 4}{4}, 0 \right) = (2, 3, 0)\). Thus, \(\overrightarrow{MN} = (2 - 1)\mathbf{i} + (3 - 4)\mathbf{j} + (0 - 2)\mathbf{k} = \mathbf{i} + \mathbf{j} - 2\mathbf{k}\).

(b) The vector equation of the line through M and N is given by \(\mathbf{r} = \overrightarrow{OM} + \lambda \overrightarrow{MN}\). Substituting the vectors, we get \(\mathbf{r} = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) + \lambda (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 2\mathbf{i} + 3\mathbf{j} + \lambda (\mathbf{i} + \mathbf{j} - 2\mathbf{k})\).

(c) To find the perpendicular distance from O to the line through M and N, consider a point P on the line: \(\overrightarrow{OP} = 2\mathbf{i} + 3\mathbf{j} + \lambda (\mathbf{i} + \mathbf{j} - 2\mathbf{k})\). The direction vector of the line is \(\mathbf{i} + \mathbf{j} - 2\mathbf{k}\). The perpendicular distance is given by the magnitude of the projection of \(\overrightarrow{OP}\) onto the normal vector to the line. Solving for \(\lambda\) when the dot product of \(\overrightarrow{OP}\) and the direction vector is zero gives \(\lambda = -\frac{5}{6}\). Substituting back, the perpendicular distance is \(\sqrt{\left( \frac{7}{6} \right)^2 + \left( \frac{13}{6} \right)^2 + \left( \frac{5}{3} \right)^2} = \sqrt{\frac{53}{6}}\).