(i) To find \(\overrightarrow{AB}\), we calculate \(\overrightarrow{OB} - \overrightarrow{OA} = (4\mathbf{i} + 6\mathbf{k}) - (\mathbf{i} + 2\mathbf{j}) = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}\).

The magnitude of \(\overrightarrow{AB}\) is \(\sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7\).

Thus, the unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}}{7}\).

(ii) The scalar product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 1 \times 4 + 2 \times 0 + 0 \times p = 4\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{1^2 + 2^2} = \sqrt{5}\) and \(|\overrightarrow{OB}| = \sqrt{4^2 + p^2} = \sqrt{16 + p^2}\).

Using the cosine formula, \(4 = \sqrt{5} \times \sqrt{16 + p^2} \times \frac{1}{5}\).

Solving for \(p\), we have \(4 = \frac{\sqrt{5} \times \sqrt{16 + p^2}}{5}\).

\(20 = \sqrt{5} \times \sqrt{16 + p^2}\).

\(400 = 5 \times (16 + p^2)\).

\(400 = 80 + 5p^2\).

\(320 = 5p^2\).

\(p^2 = 64\).

\(p = \pm 8\).

(i) For \(\overrightarrow{OA}\) to be parallel to \(\overrightarrow{OB}\), there must exist a scalar \(k\) such that:

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

Equating components, we get:

\(1 = 3k\), \(-2 = pk\), \(2 = qk\)

Solving these gives \(k = \frac{1}{3}\), \(p = -6\), \(q = 6\).

(ii) Given \(q = 2p\), the vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) are perpendicular if their dot product is zero:

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

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

\(3 + 2p = 0\)

\(p = -1.5\)

(iii) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\mathbf{i} + 1\mathbf{j} + 8\mathbf{k}) - (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})\)

\(= 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\)

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

The unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}}{7}\).

(a) To show \(OA = OB\), calculate the magnitudes:

\(|\overrightarrow{OA}| = \sqrt{(2)^2 + (-1)^2 + (0)^2} = \sqrt{4 + 1} = \sqrt{5}\)

\(|\overrightarrow{OB}| = \sqrt{(0)^2 + (1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\)

Thus, \(OA = OB = \sqrt{5}\).

To find angle \(AOB\), use the scalar product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (2)(0) + (-1)(1) + (0)(-2) = -1\)

\(\cos \theta = \frac{-1}{\sqrt{5} \times \sqrt{5}} = \frac{-1}{5}\)

\(\theta = \cos^{-1}\left(\frac{-1}{5}\right) \approx 101.5^{\circ}\)

(b) The midpoint \(M\) of \(AB\) is:

\(M = \left(\frac{2+0}{2}, \frac{-1+1}{2}, \frac{0-2}{2}\right) = (1, 0, -1)\)

Position vector of \(M\) is \(\mathbf{i} - \mathbf{k}\).

Let \(\overrightarrow{OP} = \lambda(\mathbf{i} - \mathbf{k})\).

Given \(PA : OA = \sqrt{7} : 1\), express \(AP = \sqrt{7} \cdot OA\):

\(\overrightarrow{AP} = \lambda(\mathbf{i} - \mathbf{k}) - (2\mathbf{i} - \mathbf{j})\)

\(\sqrt{(\lambda - 2)^2 + (-\lambda + 1)^2 + (-\lambda)^2} = \sqrt{7} \cdot \sqrt{5}\)

\((\lambda - 2)^2 + (-\lambda + 1)^2 + (-\lambda)^2 = 35\)

Solving gives \(\lambda = 5\) or \(\lambda = -3\).

Thus, \(\overrightarrow{OP} = 5(\mathbf{i} - \mathbf{k}) = 5\mathbf{i} - 5\mathbf{k}\) or \(-3(\mathbf{i} - \mathbf{k}) = -3\mathbf{i} + 3\mathbf{k}\).

(i) To show that \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OC}\), we calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OC} = (\mathbf{i} + 2p\mathbf{j} + q\mathbf{k}) \cdot (-(4p^2 + q^2)\mathbf{i} + 2p\mathbf{j} + q\mathbf{k})\)

\(= -(4p^2 + q^2) \cdot 1 + 2p \cdot 2p + q \cdot q\)

\(= -(4p^2 + q^2) + 4p^2 + q^2\)

\(= 0\)

Thus, \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OC}\).

(ii) The vector \(\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC}\):

\(\overrightarrow{CA} = (\mathbf{i} + 2p\mathbf{j} + q\mathbf{k}) - (-(4p^2 + q^2)\mathbf{i} + 2p\mathbf{j} + q\mathbf{k})\)

\(= (1 + 4p^2 + q^2)\mathbf{i}\)

The magnitude is:

\(|\overrightarrow{CA}| = \sqrt{(1 + 4p^2 + q^2)^2}\)

\(= \sqrt{1 + 4p^2 + q^2}\)

(iii) For \(p = 3\) and \(q = 2\), find \(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB}\):

\(\overrightarrow{BA} = (\mathbf{i} + 2(3)\mathbf{j} + 2\mathbf{k}) - (2\mathbf{j} - 6\mathbf{k})\)

\(= \mathbf{i} + 6\mathbf{j} + 2\mathbf{k} - 2\mathbf{j} + 6\mathbf{k}\)

\(= \mathbf{i} + 4\mathbf{j} + 8\mathbf{k}\)

The magnitude of \(\overrightarrow{BA}\) is:

\(\sqrt{1^2 + 4^2 + 8^2} = \sqrt{81} = 9\)

The unit vector is:

\(\frac{1}{9}(\mathbf{i} + 4\mathbf{j} + 8\mathbf{k})\)

(i) For OAB to be a straight line, \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) must be parallel. Thus, \(\begin{pmatrix} 4 \\ 2 \\ p \end{pmatrix} = k \begin{pmatrix} p \\ 1 \\ 1 \end{pmatrix}\) for some scalar \(k\). Solving, we find \(p = 2\). The unit vector in the direction of \(\overrightarrow{OA}\) is \(\frac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\).

(ii) For OA to be perpendicular to AB, \(\overrightarrow{OA} \cdot \overrightarrow{AB} = 0\). First, find \(\overrightarrow{AB} = \begin{pmatrix} 4-p \\ 1 \\ p-1 \end{pmatrix}\). Then, \(\overrightarrow{OA} \cdot \overrightarrow{AB} = (p)(4-p) + 1 + (p-1) = 0\). Simplifying gives \(5p - p^2 = 0\), so \(p = 0\) or \(p = 5\).

(iii) For OABC to be a parallelogram, \(\overrightarrow{OC} = \overrightarrow{OB} - \overrightarrow{OA}\). With \(p = 3\), \(\overrightarrow{OB} = \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix}\) and \(\overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\). Thus, \(\overrightarrow{OC} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).