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

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

Calculate the dot product: \(2(-2) + (-2)(3) + (-1)(6) = -16\)

Calculate magnitudes: \(|\overrightarrow{OA}| = \sqrt{2^2 + (-2)^2 + (-1)^2} = 3\)

\(|\overrightarrow{OB}| = \sqrt{(-2)^2 + 3^2 + 6^2} = 7\)

Substitute into the formula: \(3 \times 7 \times \cos \theta = -16\)

Solve for \(\theta\): \(\theta = 139.6^\circ\) or \(2.44^c\) or \(0.776\pi\)

(ii) Find \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ 6 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix}\)

Magnitude of \(\overrightarrow{AC} = \sqrt{0^2 + 8^2 + 6^2} = 10\)

Scale to magnitude 15: \(\frac{15}{10} \times \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 9 \end{pmatrix}\)

(iii) For \(p\overrightarrow{OA} + \overrightarrow{OC}\) to be perpendicular to \(\overrightarrow{OB}\), the dot product must be zero:

\((p\overrightarrow{OA} + \overrightarrow{OC}) \cdot \overrightarrow{OB} = 0\)

\(\begin{pmatrix} 2+2p \\ 6-2p \\ 5-p \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 3 \\ 6 \end{pmatrix} = 0\)

\(-2(2+2p) + 3(6-2p) + 6(5-p) = 0\)

Simplify: \(-4 - 4p + 18 - 6p + 30 - 6p = 0\)

\(44 - 16p = 0\)

\(p = \frac{2}{3/4}\)