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 \(3\).

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:

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

(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.

(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 \(\lambda\):

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

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

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

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