(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 \(p = -7\)