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

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 1 \cdot k + 0 \cdot (-k) + 2 \cdot 2k = k + 4k = 5k\).

For \(k = 2\), \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 10\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{5}\) and \(|\overrightarrow{OB}| = \sqrt{k^2 + (-k)^2 + (2k)^2} = \sqrt{6k^2} = \sqrt{24}\).

Using the dot product formula, \(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| \cdot |\overrightarrow{OB}| \cdot \cos \theta\).

\(10 = \sqrt{5} \cdot \sqrt{24} \cdot \cos \theta\).

\(\cos \theta = \frac{10}{\sqrt{5} \cdot \sqrt{24}}\).

\(\theta = \cos^{-1}\left(\frac{10}{\sqrt{5} \cdot \sqrt{24}}\right) \approx 24.1^\circ\).

(ii) For \(\overrightarrow{AB}\) to be a unit vector, its magnitude must be 1.

\(\overrightarrow{AB} = \begin{pmatrix} k-1 \\ -k \\ 2k-2 \end{pmatrix}\).

Magnitude: \(\sqrt{(k-1)^2 + (-k)^2 + (2k-2)^2} = 1\).

\((k-1)^2 + k^2 + (2k-2)^2 = 1\).

\((k-1)^2 = k^2 - 2k + 1\), \((2k-2)^2 = 4k^2 - 8k + 4\).

\(k^2 - 2k + 1 + k^2 + 4k^2 - 8k + 4 = 1\).

\(6k^2 - 10k + 5 = 1\).

\(6k^2 - 10k + 4 = 0\).

Solving the quadratic equation: \(k = 1\) or \(k = \frac{2}{3}\).

(i) To find \(\overrightarrow{CD}\), calculate \(3\mathbf{a} - 2\mathbf{b}\):

\(3\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} - 2\begin{pmatrix} 4 \\ 0 \\ 6 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} - \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

The magnitude of \(\overrightarrow{CD}\) is \(\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\).

Thus, the unit vector is \(\frac{1}{7} \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

(ii) The position vector of E is the midpoint of C and D:

\(\overrightarrow{OE} = \frac{1}{2}(3\mathbf{a} + 2\mathbf{b}) = \frac{1}{2}\left(\begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} + \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix}\right) = \begin{pmatrix} 7 \\ 1.5 \\ 9 \end{pmatrix}\).

Calculate \(\overrightarrow{OE} \cdot \overrightarrow{OD}\):

\(\begin{pmatrix} 7 \\ 1.5 \\ 9 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} = 56 + 0 + 108 = 164\).

The magnitudes are \(|\overrightarrow{OE}| = \sqrt{7^2 + 1.5^2 + 9^2} = \sqrt{132.25} \approx 11.5\) and \(|\overrightarrow{OD}| = \sqrt{8^2 + 0^2 + 12^2} = \sqrt{208}\).

Using the dot product formula, \(164 = 11.5 \times \sqrt{208} \times \cos \theta\).

Solving for \(\theta\), \(\theta = \cos^{-1}\left(\frac{164}{11.5 \times \sqrt{208}}\right) \approx 8.6^\circ\).

(i) To find \(\overrightarrow{AB}\), calculate \(\overrightarrow{OB} - \overrightarrow{OA}\):

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

The modulus of \(\overrightarrow{AB}\) is \(\sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = 7.\)

The unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{1}{7} \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}.\)

(ii) For \(\angle BOC = 90^\circ\), \(\overrightarrow{OB} \cdot \overrightarrow{OC} = 0.\)

Calculate the dot product:

\(\overrightarrow{OB} \cdot \overrightarrow{OC} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix} = 4 \times 1 + 2 \times 3 + (-2) \times p = 4 + 6 - 2p.\)

Set the dot product to zero:

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

\(10 = 2p\)

\(p = 5\)

(i) To find the angle between the vectors \(3\mathbf{i} - 4\mathbf{k}\) and \(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\), we use the dot product formula:

\(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\).

The dot product is \((3)(2) + (0)(3) + (-4)(-6) = 6 + 24 = 30\).

The magnitudes are \(|3\mathbf{i} - 4\mathbf{k}| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{25}\) and \(|2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49}\).

Thus, \(30 = \sqrt{25} \times \sqrt{49} \cos \theta\).

Solving for \(\theta\), we get \(\theta = \cos^{-1}\left(\frac{30}{35}\right)\), which is approximately 31° or 0.54 radians.

(ii) \(\overrightarrow{OA}\) is in the same direction as \(3\mathbf{i} - 4\mathbf{k}\) with magnitude 15:

\(\overrightarrow{OA} = (3\mathbf{i} - 4\mathbf{k}) \times \frac{15}{5} = 9\mathbf{i} - 12\mathbf{k}\).

\(\overrightarrow{OB}\) is in the same direction as \(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) with magnitude 14:

\(\overrightarrow{OB} = (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \times \frac{14}{7} = 4\mathbf{i} + 6\mathbf{j} - 12\mathbf{k}\).

(iii) To find the unit vector in the direction of \(\overrightarrow{AB}\), first find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4\mathbf{i} + 6\mathbf{j} - 12\mathbf{k}) - (9\mathbf{i} - 12\mathbf{k}) = -5\mathbf{i} + 6\mathbf{j}\).

The magnitude of \(\overrightarrow{AB}\) is \(\sqrt{(-5)^2 + 6^2} = \sqrt{61}\).

Thus, the unit vector is \(\frac{-5\mathbf{i} + 6\mathbf{j}}{\sqrt{61}}\).

(i) For u to be perpendicular to v, their dot product must be zero:

\(\mathbf{u} \cdot \mathbf{v} = p^2 \cdot 2 + (-2) \cdot (p-1) + 6 \cdot (2p+1) = 0\)

\(2p^2 - 2(p-1) + 6(2p+1) = 0\)

\(2p^2 - 2p + 2 + 12p + 6 = 0\)

\(2p^2 + 10p + 8 = 0\)

Factoring gives:

\((p+1)(p+4) = 0\)

Thus, \(p = -1\) or \(p = -4\).

(ii) For \(p = 1\), calculate the angle between u and v:

\(\mathbf{u} = \begin{pmatrix} 1 \\ -2 \\ 6 \end{pmatrix}, \mathbf{v} = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}\)

\(\mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + (-2) \cdot 0 + 6 \cdot 3 = 20\)

\(|\mathbf{u}| = \sqrt{1^2 + (-2)^2 + 6^2} = \sqrt{41}\)

\(|\mathbf{v}| = \sqrt{2^2 + 0^2 + 3^2} = \sqrt{13}\)

Using the dot product formula:

\(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} = \frac{20}{\sqrt{41} \times \sqrt{13}}\)

\(\theta = \cos^{-1}\left(\frac{20}{\sqrt{41} \times \sqrt{13}}\right) \approx 30.0^\circ\) or \(0.523 \text{ rads}\).