(i) First, find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3i + 2j + 8k) - (i - 2j + 4k) = 2i + 4j + 4k\)

\(\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = (3i + 2j + 8k) - (-i - 2j + 10k) = 4i + 4j - 2k\)

Calculate the scalar product \(\overrightarrow{AB} \cdot \overrightarrow{CB}\):

\(\overrightarrow{AB} \cdot \overrightarrow{CB} = (2i + 4j + 4k) \cdot (4i + 4j - 2k) = 2 \times 4 + 4 \times 4 + 4 \times (-2) = 8 + 16 - 8 = 16\)

Find the magnitudes of \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):

\(|\overrightarrow{AB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\)

\(|\overrightarrow{CB}| = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\)

Use the scalar product to find \(\cos \theta\):

\(\overrightarrow{AB} \cdot \overrightarrow{CB} = |\overrightarrow{AB}| \cdot |\overrightarrow{CB}| \cdot \cos \theta\)

\(16 = 6 \times 6 \times \cos \theta\)

\(\cos \theta = \frac{16}{36} = \frac{4}{9}\)

\(\theta = \cos^{-1}\left(\frac{4}{9}\right) \approx 63.6^\circ\)

(ii) Find the lengths of \(AB\), \(BC\), and \(CA\):

\(AB = |\overrightarrow{AB}| = 6\)

\(BC = |\overrightarrow{CB}| = 6\)

\(CA = |\overrightarrow{CA}| = \sqrt{(-2)^2 + 0^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40}\)

Perimeter of \(\triangle ABC = AB + BC + CA = 6 + 6 + \sqrt{40} \approx 18.32\)

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

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

\(-8 + 3 + p = 0\)

\(p = 5\)

(ii) The vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\) is given by:

\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 1 \\ p \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ p-1 \end{pmatrix}\)

The magnitude of \(\overrightarrow{AB}\) is 7:

\(\sqrt{6^2 + (-2)^2 + (p-1)^2} = 7\)

\(\sqrt{36 + 4 + (p-1)^2} = 7\)

\(36 + 4 + (p-1)^2 = 49\)

\((p-1)^2 = 9\)

\(p-1 = 3\) or \(p-1 = -3\)

\(p = 4\) or \(p = -2\)

(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:

\(\left( \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix} + p \begin{pmatrix} 0 \\ -6 \\ 8 \end{pmatrix} \right) \cdot \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} = 0\).

\(\begin{pmatrix} 2 \\ 3 - 6p \\ -6 + 8p \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} = 0\).

\(2(-2) + (3 - 6p)5 + (-6 + 8p)(-2) = 0\).

\(-4 + 15 - 30p + 12 - 16p = 0\).

\(23 - 46p = 0\).

\(p = \frac{1}{2}\).

(i) To find \(\overrightarrow{OA} \cdot \overrightarrow{OB}\), calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (2)(7) + (-8)(2) + (4)(-1) = 14 - 16 - 4 = -6\).

Since the dot product is negative, angle AOB is obtuse.

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

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (7\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (2\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}\).

Then, \(\overrightarrow{AX} = \frac{2}{5} \overrightarrow{AB} = \frac{2}{5} (5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}) = 2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\).

Now, find \(\overrightarrow{OX}\):

\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = (2\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}) + (2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\).

Finally, find the unit vector in the direction of \(\overrightarrow{OX}\):

The magnitude of \(\overrightarrow{OX}\) is \(\sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\).

Thus, the unit vector is \(\frac{1}{6} (4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})\).

(i) The vectors \(\mathbf{OA} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{OB} = 3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}\) are perpendicular if their dot product is zero:

\((2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}) = 0\)

\(2 \times 3 + 1 \times (-2) + 2 \times p = 0\)

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

\(2p = -4\)

\(p = -2\)

(ii) For \(p = 6\), the vectors are \(\mathbf{OA} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{OB} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}\). The dot product is:

\((2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) = 6 - 2 + 12 = 16\)

The magnitudes are:

\(|\mathbf{OA}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3\)

\(|\mathbf{OB}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{49} = 7\)

Using the cosine formula:

\(16 = 3 \times 7 \times \cos \theta\)

\(\cos \theta = \frac{16}{21}\)

\(\theta \approx 40^\circ\)

(iii) The vector \(\mathbf{AB} = (3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}) - (2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = \mathbf{i} - 3\mathbf{j} + (p - 2)\mathbf{k}\).

The length of \(\mathbf{AB}\) is 3.5:

\(\sqrt{1^2 + (-3)^2 + (p-2)^2} = 3.5\)

\(1 + 9 + (p-2)^2 = 12.25\)

\((p-2)^2 = 2.25\)

\(p-2 = \pm 1.5\)

\(p = 3.5\) or \(p = 0.5\)