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

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos AOB\)

Given \(p = 2\), \(\overrightarrow{OA} = 3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\) and \(\overrightarrow{OB} = 6\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}\).

Calculate the dot product:

\(3 \times 6 + 2 \times 6 + (-4) \times 3 = 18\)

Calculate magnitudes:

\(|\overrightarrow{OA}| = \sqrt{3^2 + 2^2 + (-4)^2} = \sqrt{29}\)

\(|\overrightarrow{OB}| = \sqrt{6^2 + 6^2 + 3^2} = 9\)

Substitute into the formula:

\(\sqrt{29} \times 9 \times \cos AOB = 18\)

\(\cos AOB = \frac{18}{\sqrt{29} \times 9}\)

\(AOB = \cos^{-1}\left(\frac{18}{\sqrt{29} \times 9}\right) \approx 68.2^\circ\) or \(1.19 \text{ radians}\)

(ii) For \(\overrightarrow{AB}\) to be parallel to \(\overrightarrow{OC}\), \(\overrightarrow{AB} = \lambda \overrightarrow{OC}\) for some scalar \(\lambda\).

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

\(\overrightarrow{OC} = (p - 1)\mathbf{i} + 2\mathbf{j} + q\mathbf{k}\)

Comparing the \(\mathbf{j}\) components:

\(4 = 2 \lambda \Rightarrow \lambda = 2\)

Comparing the \(\mathbf{i}\) components:

\(3 = 2(p - 1) \Rightarrow 3 = 2p - 2 \Rightarrow 2p = 5 \Rightarrow p = 2.5\)

Comparing the \(\mathbf{k}\) components:

\(3 + 2p = 2q \Rightarrow 3 + 2(2.5) = 2q \Rightarrow 8 = 2q \Rightarrow q = 4\)

Thus, \(p = 2.5\) and \(q = 4\).