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

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