(i) To find \(\overrightarrow{CP}\), note that \(C\) is at \((6, 0, 2)\) and \(P\) is at \((0, 6, 0)\). Thus, \(\overrightarrow{CP} = (0 - 6)\mathbf{i} + (6 - 0)\mathbf{j} + (0 - 2)\mathbf{k} = -6\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\).

For \(\overrightarrow{CQ}\), \(Q\) is at \((0, 6, 5)\). Thus, \(\overrightarrow{CQ} = (0 - 6)\mathbf{i} + (6 - 0)\mathbf{j} + (5 - 2)\mathbf{k} = -6\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}\).

(ii) The scalar product \(\overrightarrow{CP} \cdot \overrightarrow{CQ} = (-6)(-6) + (6)(6) + (-2)(3) = 36 + 36 - 6 = 66\).

The magnitudes are \(|\overrightarrow{CP}| = \sqrt{(-6)^2 + 6^2 + (-2)^2} = \sqrt{76}\) and \(|\overrightarrow{CQ}| = \sqrt{(-6)^2 + 6^2 + 3^2} = \sqrt{81}\).

Using the formula for the angle, \(66 = \sqrt{76} \times \sqrt{81} \times \cos \theta\), we find \(\cos \theta = \frac{66}{\sqrt{76} \times \sqrt{81}}\).

Thus, \(\theta = \cos^{-1}\left(\frac{66}{\sqrt{76} \times \sqrt{81}}\right) \approx 32.7^{\circ}\) or \(0.571 \text{ rad}\).

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

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (-6) + 2 + 12 = 8\)

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{14}\) and \(|\overrightarrow{OB}| = \sqrt{29}\).

\(\cos AOB = \frac{8}{\sqrt{14} \sqrt{29}}\)

\(AOB = \cos^{-1}\left(\frac{8}{\sqrt{14} \sqrt{29}}\right) \approx 66.6^\circ\)

(ii) \(\overrightarrow{BC} = \overrightarrow{OB} - \overrightarrow{OC}\)

\(\overrightarrow{OC} = p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)

\(\overrightarrow{BC} = (-3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}) - p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)

\(\overrightarrow{BC} = 3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)

(iii) For \(BC\) to be perpendicular to \(OA\), \(\overrightarrow{BC} \cdot \overrightarrow{OA} = 0\).

\((3 + 2p) \cdot 2 + (p - 2) \cdot 1 + (4 - 3p) \cdot (-3) = 0\)

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

\(6 + 4p + p - 2 - 12 + 9p = 0\)

\(14p - 8 = 0\)

\(p = \frac{4}{7}\) or 0.571

(i) To find \(a\), we use the similarity of triangles. The triangle \(OPD\) is similar to \(OPA\). The ratio of the sides gives:

\(\frac{10 - a}{10} = \frac{6}{10}\)

Solving for \(a\):

\(10 - a = 6\)

\(a = 4\)

(ii) To express \(\overrightarrow{BG}\), we find the coordinates of \(B\) and \(G\). \(B\) is at \((10, 10, 0)\) and \(G\) is at \((0, 6, 4)\). Thus,

\(\overrightarrow{BG} = (0 - 10)\mathbf{i} + (6 - 10)\mathbf{j} + (4 - 0)\mathbf{k}\)

\(\overrightarrow{BG} = -10\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\)

(iii) To find angle \(GBA\), we use the scalar product:

\(\overrightarrow{BG} \cdot \overrightarrow{BA} = 40\)

\(\cos GBA = \frac{40}{\sqrt{132} \times \sqrt{100}}\)

\(GBA = \cos^{-1}\left(\frac{40}{\sqrt{13200}}\right)\)

\(GBA = 69.6^\circ\)

First, determine the vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\):

\(\overrightarrow{AC} = -6\mathbf{i} + 10\mathbf{k}\)

\(\overrightarrow{BC} = -8\mathbf{j} + 10\mathbf{k}\)

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

\(\overrightarrow{AC} \cdot \overrightarrow{BC} = (-6)(0) + (0)(-8) + (10)(10) = 100\)

Find the magnitudes of \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\):

\(|\overrightarrow{AC}| = \sqrt{(-6)^2 + 0^2 + 10^2} = \sqrt{136}\)

\(|\overrightarrow{BC}| = \sqrt{0^2 + (-8)^2 + 10^2} = \sqrt{164}\)

Use the scalar product formula to find \(\cos \angle ACB\):

\(\overrightarrow{AC} \cdot \overrightarrow{BC} = |\overrightarrow{AC}| |\overrightarrow{BC}| \cos \angle ACB\)

\(100 = \sqrt{136} \sqrt{164} \cos \angle ACB\)

\(\cos \angle ACB = \frac{100}{\sqrt{136} \sqrt{164}}\)

Calculate \(\angle ACB\):

\(\angle ACB = \cos^{-1}\left(\frac{100}{\sqrt{136} \sqrt{164}}\right) \approx 48.0^{\circ}\)