(i) First, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}\).

Then, \(\overrightarrow{AC} = 2\overrightarrow{AB} = 2 \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}\).

Now, \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + \begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{7} \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\).

(ii) We have \(\overrightarrow{OD} = m\overrightarrow{OA} + n\overrightarrow{OB}\).

\(m \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + n \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ k \end{pmatrix}\).

This gives the equations:

\(4m + 3n = 1\)

\(m + 2n = 4\)

Solving these simultaneously, we find \(m = -2\) and \(n = 3\).

Substitute \(m\) and \(n\) into the third component equation:

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

\(k = -8\).

(i) To find the angle \(AOB\), we use the dot product formula:

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

Given \(\mathbf{a} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}\), the dot product is:

\(\mathbf{a} \cdot \mathbf{b} = (-3)(-1) + (6)(2) + (3)(4) = 3 + 12 + 12 = 27\)

The magnitudes are:

\(|\mathbf{a}| = \sqrt{(-3)^2 + 6^2 + 3^2} = \sqrt{54}\)

\(|\mathbf{b}| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}\)

Thus, \(27 = \sqrt{54} \times \sqrt{21} \times \cos \theta\)

\(\cos \theta = \frac{27}{\sqrt{54} \times \sqrt{21}}\)

\(\theta = \cos^{-1}\left(\frac{27}{\sqrt{54} \times \sqrt{21}}\right) \approx 36.7^\circ\) or \(0.641\) radians

(ii) To find \(\overrightarrow{OC}\), first find \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}\).

Then, \(\overrightarrow{AC} = 3\overrightarrow{AB} = 3 \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix}\).

\(\overrightarrow{OC} = \mathbf{a} + \overrightarrow{AC} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix}\).

The magnitude of \(\overrightarrow{OC}\) is \(\sqrt{3^2 + (-6)^2 + 6^2} = 9\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{9} \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3} \end{pmatrix}\).

Thus, the unit vector is \(\frac{1}{9} \mathbf{i} - \frac{2}{3} \mathbf{j} + \frac{2}{3} \mathbf{k}\).

(i) To find \(\cos POQ\), we use the scalar product formula:

\(\overrightarrow{OP} \cdot \overrightarrow{OQ} = (-2)(2) + (3)(1) + (1)(3) = -4 + 3 + 3 = 2\).

The magnitudes are \(|\overrightarrow{OP}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14}\) and \(|\overrightarrow{OQ}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).

Thus, \(\cos \theta = \frac{\overrightarrow{OP} \cdot \overrightarrow{OQ}}{|\overrightarrow{OP}| |\overrightarrow{OQ}|} = \frac{2}{\sqrt{14} \times \sqrt{14}} = \frac{2}{14} = \frac{1}{7}\).

(ii) The vector \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \\ q \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ q-1 \end{pmatrix}\).

The length of \(\overrightarrow{PQ}\) is given by \(\sqrt{4^2 + (-2)^2 + (q-1)^2} = 6\).

Squaring both sides, we have \(16 + 4 + (q-1)^2 = 36\).

Thus, \((q-1)^2 = 16\), giving \(q-1 = 4\) or \(q-1 = -4\).

Therefore, \(q = 5\) or \(q = -3\).

(i) The scalar product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 8 - 9 - 2 = -3\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{14}\) and \(|\overrightarrow{OB}| = \sqrt{29}\).

Using the formula \(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos \theta\), we have:

\(-3 = \sqrt{14} \times \sqrt{29} \times \cos \theta\)

\(\cos \theta = \frac{-3}{\sqrt{14} \times \sqrt{29}}\)

\(\theta = \cos^{-1}\left(\frac{-3}{\sqrt{14} \times \sqrt{29}}\right) \approx 99^\circ\).

(ii) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4 - 2)\mathbf{i} + (-3 - 3)\mathbf{j} + (2 + 1)\mathbf{k} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}\).

The magnitude of \(\overrightarrow{AB}\) is \(\sqrt{2^2 + (-6)^2 + 3^2} = \sqrt{49} = 7\).

The unit vector is \(\frac{1}{7}(2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k})\).

(iii) \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = -2\mathbf{i} + 3\mathbf{j} + (p + 1)\mathbf{k}\).

Given \(|\overrightarrow{AB}| = |\overrightarrow{AC}|\), we have:

\(49 = 4 + 9 + (p + 1)^2\)

\(49 = 13 + (p + 1)^2\)

\((p + 1)^2 = 36\)

\(p + 1 = 6\) or \(p + 1 = -6\)

\(p = 5\) or \(p = -7\)

(i) To find \(\overrightarrow{AX}\), calculate \(\overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix}.\)

To show AXB is a straight line, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 10 \\ 2 \\ 11 \end{pmatrix} - \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 18 \\ 6 \\ 9 \end{pmatrix}.\)

Since \(\overrightarrow{AB} = 3 \overrightarrow{AX}\), AXB is a straight line.

(ii) To show CX is perpendicular to AX, find \(\overrightarrow{CX} = \overrightarrow{OX} - \overrightarrow{OC} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -8 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix}.\)

Calculate the dot product \(\overrightarrow{CX} \cdot \overrightarrow{AX} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix} = -18 + 12 + 6 = 0.\)

Since the dot product is zero, CX is perpendicular to AX.

(iii) To find the area of triangle ABC, calculate the magnitudes \(|\overrightarrow{CX}| = \sqrt{(-3)^2 + 6^2 + 2^2} = 7\) and \(|\overrightarrow{AB}| = \sqrt{18^2 + 6^2 + 9^2} = 21.\)

The area is \(\frac{1}{2} \times |\overrightarrow{CX}| \times |\overrightarrow{AB}| = \frac{1}{2} \times 7 \times 21 = 73.5.\)