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:

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