(i) To show that \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OC}\), we calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OC} = (\mathbf{i} + 2p\mathbf{j} + q\mathbf{k}) \cdot (-(4p^2 + q^2)\mathbf{i} + 2p\mathbf{j} + q\mathbf{k})\)

\(= -(4p^2 + q^2) \cdot 1 + 2p \cdot 2p + q \cdot q\)

\(= -(4p^2 + q^2) + 4p^2 + q^2\)

\(= 0\)

Thus, \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OC}\).

(ii) The vector \(\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC}\):

\(\overrightarrow{CA} = (\mathbf{i} + 2p\mathbf{j} + q\mathbf{k}) - (-(4p^2 + q^2)\mathbf{i} + 2p\mathbf{j} + q\mathbf{k})\)

\(= (1 + 4p^2 + q^2)\mathbf{i}\)

The magnitude is:

\(|\overrightarrow{CA}| = \sqrt{(1 + 4p^2 + q^2)^2}\)

\(= \sqrt{1 + 4p^2 + q^2}\)

(iii) For \(p = 3\) and \(q = 2\), find \(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB}\):

\(\overrightarrow{BA} = (\mathbf{i} + 2(3)\mathbf{j} + 2\mathbf{k}) - (2\mathbf{j} - 6\mathbf{k})\)

\(= \mathbf{i} + 6\mathbf{j} + 2\mathbf{k} - 2\mathbf{j} + 6\mathbf{k}\)

\(= \mathbf{i} + 4\mathbf{j} + 8\mathbf{k}\)

The magnitude of \(\overrightarrow{BA}\) is:

\(\sqrt{1^2 + 4^2 + 8^2} = \sqrt{81} = 9\)

The unit vector is:

\(\frac{1}{9}(\mathbf{i} + 4\mathbf{j} + 8\mathbf{k})\)