(i) The position vectors are \(\mathbf{OA} = \mathbf{i} + 7\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{OB} = -5\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}\).

The dot product \(\mathbf{OA} \cdot \mathbf{OB} = (1)(-5) + (7)(5) + (2)(6) = -5 + 35 + 12 = 42\).

The magnitudes are \(|\mathbf{OA}| = \sqrt{1^2 + 7^2 + 2^2} = \sqrt{54}\) and \(|\mathbf{OB}| = \sqrt{(-5)^2 + 5^2 + 6^2} = \sqrt{86}\).

Using the dot product formula: \(\mathbf{OA} \cdot \mathbf{OB} = |\mathbf{OA}| |\mathbf{OB}| \cos \theta\), we have:

\(42 = \sqrt{54} \sqrt{86} \cos \theta\).

Solving for \(\cos \theta\), we get \(\cos \theta = \frac{42}{\sqrt{54} \sqrt{86}}\).

Thus, \(\theta = \cos^{-1}\left(\frac{42}{\sqrt{54} \sqrt{86}}\right) \approx 0.907\) radians.

(ii) Given \(\overrightarrow{AB} = 2\overrightarrow{BC}\), we find \(\overrightarrow{BC} = \frac{1}{2} \overrightarrow{AB}\).

\(\overrightarrow{AB} = \mathbf{OB} - \mathbf{OA} = (-5\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}) - (\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}) = -6\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\).

Thus, \(\overrightarrow{BC} = \frac{1}{2}(-6\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) = -3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\).

\(\overrightarrow{OC} = \mathbf{OB} + \overrightarrow{BC} = (-5\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}) + (-3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = -8\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}\).

The magnitude of \(\overrightarrow{OC}\) is \(\sqrt{(-8)^2 + 4^2 + 8^2} = \sqrt{144} = 12\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{-8\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}}{12}\).

(i) The vector \(\overrightarrow{AB}\) is given by \(\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\).

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

Thus, the unit vector in the direction of \(\overrightarrow{AB}\) is \(\pm \frac{1}{6} \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\).

(ii) For \(\angle AOC = 90^\circ\), the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OC} = 0\).

\(\overrightarrow{OA} \cdot \overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 2 \\ p \end{pmatrix} = 4 + 6 - p = 0\).

Solving for \(p\), we get \(p = 10\).

(iii) The length of \(\overrightarrow{AD}\) is given by \(\sqrt{(-2)^2 + 3^2 + (q+1)^2} = 7\).

\((-2)^2 + 3^2 + (q+1)^2 = 49\).

\(4 + 9 + (q+1)^2 = 49\).

\((q+1)^2 = 36\).

Solving \((q+1)^2 = 36\) gives \(q+1 = 6\) or \(q+1 = -6\).

Thus, \(q = 5\) or \(q = -7\).

(i) Calculate \(\overrightarrow{BA} = \boldsymbol{a} - \boldsymbol{b} = \boldsymbol{i} + 2\boldsymbol{j} - 3\boldsymbol{k}\).

Calculate \(\overrightarrow{BC} = \boldsymbol{c} - \boldsymbol{b} = -2\boldsymbol{i} + 4\boldsymbol{j} + 2\boldsymbol{k}\).

Find the dot product: \((-2) + 8 - 6 = 0\).

Since the dot product is zero, \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are perpendicular.

(ii) Calculate \(\overrightarrow{BC} = \boldsymbol{c} - \boldsymbol{b} = -2\boldsymbol{i} + 4\boldsymbol{j} + 2\boldsymbol{k}\).

Calculate \(\overrightarrow{AD} = \boldsymbol{d} - \boldsymbol{a} = -5\boldsymbol{i} + 10\boldsymbol{j} + 5\boldsymbol{k}\).

These vectors are in the same ratio, hence they are parallel.

Find the ratio of lengths: \(\text{Ratio} = 2:5\) (or \(\sqrt{24} : \sqrt{150}\)).

(i) To find the angle \(\theta\) between \(\mathbf{a}\) and \(\mathbf{b}\), use the dot product formula:

\(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = 2 \times 2 + (-2) \times 6 + 1 \times 3 = 4 - 12 + 3 = -5\).

The magnitudes are \(|\mathbf{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\) and \(|\mathbf{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7\).

Using \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\), we have:

\(-5 = 3 \times 7 \times \cos \theta\).

\(\cos \theta = -\frac{5}{21}\).

\(\theta = \cos^{-1}\left(-\frac{5}{21}\right) \approx 103.8^\circ\) or \(1.81\) radians.

(ii) \(\mathbf{b}\) and \(\mathbf{c}\) are perpendicular if \(\mathbf{b} \cdot \mathbf{c} = 0\).

\(\mathbf{b} \cdot \mathbf{c} = 2p + 6p + 3(p+1) = 11p + 3\).

Set \(11p + 3 = 0\) to find \(p\).

\(11p = -3\).

\(p = -\frac{3}{11}\).

(i) To find the value of \(k\) for which angle \(AOB\) is \(90^\circ\), we use the dot product formula:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

\(\begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ k \\ -3 \end{pmatrix} = 6 \times 3 + (-2) \times k + (-6) \times (-3) = 0\).

\(18 - 2k + 18 = 0\).

\(36 - 2k = 0\).

\(k = 18\).

(ii) To find the values of \(k\) for which the lengths of \(OA\) and \(OB\) are equal, we equate their magnitudes:

\(\sqrt{6^2 + (-2)^2 + (-6)^2} = \sqrt{3^2 + k^2 + (-3)^2}\).

\(\sqrt{36 + 4 + 36} = \sqrt{9 + k^2 + 9}\).

\(\sqrt{76} = \sqrt{k^2 + 18}\).

\(76 = k^2 + 18\).

\(k^2 = 58\).

\(k = \pm \sqrt{58}\) or \(k \approx \pm 7.62\).

(iii) For \(k = 4\), find the unit vector in the direction of \(\overrightarrow{OC}\).

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

\(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix} + \begin{pmatrix} -2 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix}\).

The unit vector is \(\frac{1}{6} \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix}\) or \(\frac{1}{6} (4\mathbf{i} + 2\mathbf{j} - 4\mathbf{k})\).