(i) For \(\mathbf{u}\) to be perpendicular to \(\mathbf{v}\), their dot product must be zero:

\(\mathbf{u} \cdot \mathbf{v} = q \cdot 8 + 2 \cdot (q-1) + 6 \cdot (q^2-7) = 0\)

\(8q + 2q - 2 + 6q^2 - 42 = 0\)

\(6q^2 + 10q - 44 = 0\)

Solving this quadratic equation gives \(q = 2\) and \(q = -\frac{11}{3}\).

(ii) When \(q = 0\), \(\mathbf{u} = \begin{pmatrix} 0 \\ 2 \\ 6 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 8 \\ -1 \\ -7 \end{pmatrix}\).

The dot product \(\mathbf{u} \cdot \mathbf{v} = 0 \cdot 8 + 2 \cdot (-1) + 6 \cdot (-7) = -2 - 42 = -44\).

The magnitudes are \(|\mathbf{u}| = \sqrt{0^2 + 2^2 + 6^2} = \sqrt{40}\) and \(|\mathbf{v}| = \sqrt{8^2 + (-1)^2 + (-7)^2} = \sqrt{114}\).

\(\cos \theta = \frac{-44}{\sqrt{40} \times \sqrt{114}} = \frac{-44}{\sqrt{4560}} = \frac{-4}{\sqrt{11}}\).

\(\theta = \cos^{-1}\left(\frac{-4}{\sqrt{11}}\right) \approx 130.7^\circ\) or \(2.28 \text{ radians}\).

(i) To find \(\overrightarrow{AC}\), use the formula \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}\).

\(\overrightarrow{AC} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix}\).

(ii) The mid-point M of AC is given by \(\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC})\).

\(\overrightarrow{OM} = \frac{1}{2} \left( \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} \right) = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}\).

The unit vector in the direction of \(\overrightarrow{OM}\) is \(\frac{1}{\sqrt{5}} \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}\).

(iii) To find \(\overrightarrow{AB}\), use \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\).

\(\overrightarrow{AB} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}\).

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix} = -4 + 24 - 12 = 8\).

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{(-2)^2 + 6^2 + 3^2} = \sqrt{49} = 7\) and \(|\overrightarrow{AC}| = \sqrt{2^2 + 4^2 + (-4)^2} = \sqrt{36} = 6\).

Using the dot product formula, \(\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}| |\overrightarrow{AC}| \cos \theta\), we have \(8 = 7 \times 6 \times \cos \theta\).

Solving for \(\theta\), \(\cos \theta = \frac{8}{42} = \frac{4}{21}\).

\(\theta \approx 79.0^\circ\).

(i) To find \(\overrightarrow{AB}\), calculate \(\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -10 \\ 3 \\ -13 \end{pmatrix} - \begin{pmatrix} 8 \\ -6 \\ 5 \end{pmatrix} = \begin{pmatrix} -18 \\ 9 \\ -18 \end{pmatrix}\).

The magnitude \(|\overrightarrow{AB}| = \sqrt{(-18)^2 + 9^2 + (-18)^2} = 27\).

To find \(\overrightarrow{BC}\), calculate \(\overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} - \begin{pmatrix} -10 \\ 3 \\ -13 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \\ 12 \end{pmatrix}\).

The magnitude \(|\overrightarrow{BC}| = \sqrt{12^2 + (-6)^2 + 12^2} = 18\).

Since \(|\overrightarrow{AB}|, |\overrightarrow{BC}|, |\overrightarrow{CD}|\) form a geometric progression, \(|\overrightarrow{CD}| = \frac{18}{27} \times 18 = 12\).

(ii) Since \(D\) lies on the line through \(B\) and \(C\), \(\overrightarrow{OD} = \overrightarrow{OB} + t \cdot \overrightarrow{BC}\).

Using the geometric progression, \(\overrightarrow{CD} = \pm \frac{18}{27} \times \overrightarrow{BC}\).

Calculate \(\overrightarrow{OD} = \begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} \pm \frac{18}{27} \times \begin{pmatrix} 12 \\ -6 \\ 12 \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \\ 8 \end{pmatrix}\) or \(\begin{pmatrix} -7 \\ 1 \\ -9 \end{pmatrix}\).