(i) To find when \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OB}\), their dot product must be zero:

\((1)(-7) + (3)(1-p) + (p)(p) = 0\)

\(-7 + 3 - 3p + p^2 = 0\)

\(p^2 - 3p - 4 = 0\)

Factoring gives \((p+1)(p-4) = 0\), so \(p = -1\) or \(p = 4\).

(ii) The magnitudes are \(a = \sqrt{1^2 + 3^2 + p^2}\) and \(b = \sqrt{(-7)^2 + (1-p)^2 + p^2}\).

Given \(b^2 = 2a^2\):

\(49 + (1-p)^2 + p^2 = 2(1 + 9 + p^2)\)

\(49 + 1 - 2p + p^2 + p^2 = 2(10 + p^2)\)

\(50 - 2p + 2p^2 = 20 + 2p^2\)

\(50 - 2p = 20\)

\(2p = 30\)

\(p = 15\).

(iii) When \(p = -8\), \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (-7\mathbf{i} + 9\mathbf{j} + (-8)\mathbf{k}) - (\mathbf{i} + 3\mathbf{j} - 8\mathbf{k})\).

\(\overrightarrow{AB} = -8\mathbf{i} + 6\mathbf{j}\).

The magnitude \(|\overrightarrow{AB}| = \sqrt{(-8)^2 + 6^2} = 10\).

The unit vector is \(\frac{1}{10}(-8\mathbf{i} + 6\mathbf{j})\).