(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 \(a = \frac{49}{23}\).