(i) The dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB}\) is calculated as:

\((4\mathbf{i} + 7\mathbf{j} - p\mathbf{k}) \cdot (8\mathbf{i} - \mathbf{j} - p\mathbf{k}) = 4 \times 8 + 7 \times (-1) + (-p) \times (-p)\)

\(= 32 - 7 + p^2 = 25 + p^2\)

(ii) For \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) to be perpendicular, their dot product must be zero:

\(25 + p^2 = 0\)

This equation has no real solutions since \(p^2\) cannot be negative.

(iii) For angle AOB to be 60°, we use the cosine formula:

\(\cos 60^{\circ} = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}| |\overrightarrow{OB}|}\)

\(|\overrightarrow{OA}| = \sqrt{4^2 + 7^2 + (-p)^2} = \sqrt{65 + p^2}\)

\(|\overrightarrow{OB}| = \sqrt{8^2 + (-1)^2 + (-p)^2} = \sqrt{65 + p^2}\)

\(\frac{25 + p^2}{65 + p^2} = \frac{1}{2}\)

Solving \(25 + p^2 = \frac{1}{2}(65 + p^2)\) gives:

\(50 + 2p^2 = 65 + p^2\)

\(p^2 = 15\)

\(p = \pm \sqrt{15}\) or \(p = \pm 3.87\)