(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 \(\mathbf{A'} = \mathbf{OP} + (\mathbf{OP} - \mathbf{OA})\).

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

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