(i) To find the angle \(AOB\), we use the scalar product formula:
\(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos \theta\)
Calculate \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 2 \times 0 + 3 \times (-6) + (-6) \times 8 = -18 - 48 = -66\).
Calculate magnitudes: \(|\overrightarrow{OA}| = \sqrt{2^2 + 3^2 + (-6)^2} = 7\) and \(|\overrightarrow{OB}| = \sqrt{0^2 + (-6)^2 + 8^2} = 10\).
Substitute into the formula: \(-66 = 7 \times 10 \times \cos \theta\).
\(\cos \theta = \frac{-66}{70}\).
\(\theta = \cos^{-1}\left(\frac{-66}{70}\right) = 160.5^\circ\).
(ii) Find \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ 4 \end{pmatrix}\).
Magnitude of \(\overrightarrow{AC} = \sqrt{(-4)^2 + 2^2 + 4^2} = 6\).
To find a vector in the same direction with magnitude 30, scale \(\overrightarrow{AC}\) by \(\frac{30}{6} = 5\).
Vector = \(5 \times \begin{pmatrix} -4 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} -20 \\ 10 \\ 20 \end{pmatrix}\).
(iii) For \(\overrightarrow{OA} + p \overrightarrow{OB}\) to be perpendicular to \(\overrightarrow{OC}\), their dot product must be zero:
\(\left( \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix} + p \begin{pmatrix} 0 \\ -6 \\ 8 \end{pmatrix} \right) \cdot \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} = 0\).
\(\begin{pmatrix} 2 \\ 3 - 6p \\ -6 + 8p \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} = 0\).
\(2(-2) + (3 - 6p)5 + (-6 + 8p)(-2) = 0\).
\(-4 + 15 - 30p + 12 - 16p = 0\).
\(23 - 46p = 0\).
\(p = \frac{1}{2}\).
