(i) To find the value of \(p\) when \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OB}\), set the scalar product to zero:

\((p-6)(4-2p) + (2p-6)p + 1 \times 2 = 0\)

\(-2p^2 + 16p - 24 + 2p^2 - 6p + 2 = 0\)

\(10p - 22 = 0\)

Solving for \(p\), we get \(p = 2.2\).

(ii) For \(OAB\) to be a straight line, \(\overrightarrow{OB} = 2 \overrightarrow{OA}\):

\(4-2p = 2(p-6)\) and \(p = 2(2p-6)\)

Solving \(4-2p = 2(p-6)\), we get \(p = 4\).

Thus, \(\overrightarrow{OA} = \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} -4 \\ 4 \\ 2 \end{pmatrix}\).

The length of \(OA\) is \(\sqrt{(-2)^2 + 2^2 + 1^2} = 3\).

(a) To find the cosine of angle BAC, we first find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ 3 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix}\)

The scalar product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 2 \times 4 + (-2) \times 0 + 1 \times (-3) = 8 - 3 = 5\).

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\) and \(|\overrightarrow{AC}| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{25} = 5\).

Thus, \(\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{5}{3 \times 5} = \frac{1}{3}\).

(b) To find the area of triangle ABC, we use the formula \(\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|\).

The cross product \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ 4 & 0 & -3 \end{vmatrix} = \mathbf{i}((-2)(-3) - 1(0)) - \mathbf{j}(2(-3) - 1(4)) + \mathbf{k}(2(0) - (-2)(4))\).

\(= \mathbf{i}(6) - \mathbf{j}(-6 - 4) + \mathbf{k}(8) = 6\mathbf{i} + 10\mathbf{j} + 8\mathbf{k}\).

The magnitude \(|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{6^2 + 10^2 + 8^2} = \sqrt{36 + 100 + 64} = \sqrt{200} = 10\sqrt{2}\).

Thus, the area is \(\frac{1}{2} \times 10\sqrt{2} = 5\sqrt{2}\).

(i) For \(ABC\) to be a straight line, vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) must be parallel. Calculate \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 1 \\ p-5 \\ q+2 \end{pmatrix}\). Equating the ratios gives:

\(\frac{2}{1} = \frac{3}{p-5} = \frac{1}{q+2}\)

Solving these gives \(p = 6.5\) and \(q = -1.5\).

(ii) For \(\angle BAC = 90^\circ\), vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) must be perpendicular. Calculate \(\overrightarrow{AC} = \begin{pmatrix} 3 \\ p-2 \\ q+3 \end{pmatrix}\). The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 0\) gives:

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

Solving gives \(q = -3p - 3\).

(iii) For \(p = 3\) and \(AB = AC\), equate the magnitudes:

\(AB^2 = 4 + 9 + 1 = 14\)

\(AC^2 = 9 + 1 + (q+3)^2\)

Equating gives \((q+3)^2 = 4\), so \(q+3 = \pm 2\).

Thus, \(q = -1\) or \(q = -5\).

(i) To show that \(\angle ABC = 90^\circ\), we need to show that vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are perpendicular.

First, find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} - \begin{pmatrix} 3 \\ 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}.\)

Next, find \(\overrightarrow{BC}\):

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 6 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}.\)

Calculate the dot product \(\overrightarrow{AB} \cdot \overrightarrow{BC}\):

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 2 \times 1 + (-3) \times 2 + 1 \times 4 = 2 - 6 + 4 = 0.\)

Since the dot product is zero, \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are perpendicular, so \(\angle ABC = 90^\circ\).

(ii) To find the area of triangle \(ABC\), use the formula:

\(\text{Area} = \frac{1}{2} \times |\overrightarrow{AB}| \times |\overrightarrow{BC}|.\)

Calculate the magnitudes:

\(|\overrightarrow{AB}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}.\)

\(|\overrightarrow{BC}| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21}.\)

Thus, the area is:

\(\text{Area} = \frac{1}{2} \times \sqrt{14} \times \sqrt{21} = \frac{1}{2} \times \sqrt{294}.\)

\(\sqrt{294} \approx 17.146,\) so the area is approximately \(8.6\) to 1 decimal place.

(i) Calculate the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 6 + 4 + 16 = 26\).

Find the magnitudes: \(|\overrightarrow{OA}| = \sqrt{36} = 6\) and \(|\overrightarrow{OB}| = \sqrt{26}\).

Use the cosine formula: \(\cos AOB = \frac{26}{6\sqrt{26}}\).

Calculate \(AOB\): \(AOB = \cos^{-1}\left(\frac{26}{6\sqrt{26}}\right) \approx 31.8^{\circ}\).

(ii) Find \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}\).

Calculate \(\overrightarrow{OC} = \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix} + 2\begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}\).

Find the unit vector: \(\frac{1}{6} \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}\).

(iii) Calculate \(|\overrightarrow{OC}| = \sqrt{16 + 4 + 16} = 6\).

Since \(|\overrightarrow{OC}| = |\overrightarrow{OA}| = 6\), triangle OAC is isosceles.