(a) To find \(\overrightarrow{OD}\), use the property of a parallelogram: \(\overrightarrow{AB} = \overrightarrow{CD}\). Calculate \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\).

Then, \(\overrightarrow{OD} = \overrightarrow{OC} + \overrightarrow{AB} = \begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ -4 \\ -3 \end{pmatrix}\).

(b) To find \(\cos \theta\), use the dot product formula: \(\overrightarrow{BA} \cdot \overrightarrow{BC} = \|\overrightarrow{BA}\| \|\overrightarrow{BC}\| \cos \theta\).

Calculate \(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}\).

Calculate \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ -5 \\ -6 \end{pmatrix}\).

Dot product: \(\overrightarrow{BA} \cdot \overrightarrow{BC} = (-2)(-1) + (-2)(-5) + (1)(-6) = 2 + 10 - 6 = 6\).

Magnitudes: \(\|\overrightarrow{BA}\| = \sqrt{(-2)^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\), \(\|\overrightarrow{BC}\| = \sqrt{(-1)^2 + (-5)^2 + (-6)^2} = \sqrt{62}\).

\(\cos \theta = \frac{6}{3 \times \sqrt{62}} = \frac{2}{\sqrt{62}}\).

(c) The area of parallelogram \(ABCD\) is given by \(\text{Area} = \|\overrightarrow{BA}\| \|\overrightarrow{BC}\| \sin \theta\).

\(\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{2}{\sqrt{62}}\right)^2} = \sqrt{\frac{58}{62}}\).

\(\text{Area} = 3 \times \sqrt{62} \times \sqrt{\frac{58}{62}} = 3 \sqrt{58}\).

(a) Since Q is the midpoint of PR, we have:

\(\mathbf{q} = \frac{\mathbf{p} + \mathbf{r}}{2}\)

Solving for \(\mathbf{r}\):

\(2\mathbf{q} = \mathbf{p} + \mathbf{r}\)

\(\mathbf{r} = 2\mathbf{q} - \mathbf{p}\)

(b) The magnitude of the vector \(6\mathbf{i} + a\mathbf{j} + b\mathbf{k}\) is 21, so:

\(6^2 + a^2 + b^2 = 21^2\)

\(36 + a^2 + b^2 = 441\)

\(a^2 + b^2 = 405\)

The vector is perpendicular to \(3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), so:

\(6 \times 3 + a \times 2 + b \times 2 = 0\)

\(18 + 2a + 2b = 0\)

\(2a + 2b = -18\)

\(a + b = -9\)

Substitute \(b = -9 - a\) into \(a^2 + b^2 = 405\):

\(a^2 + (-9-a)^2 = 405\)

\(a^2 + (81 + 18a + a^2) = 405\)

\(2a^2 + 18a + 81 = 405\)

\(2a^2 + 18a - 324 = 0\)

\(a^2 + 9a - 162 = 0\)

Solving the quadratic equation:

\(a = \frac{-9 \pm \sqrt{9^2 + 4 \times 162}}{2}\)

\(a = 9 \text{ or } a = -18\)

Thus, \(b = -18 \text{ or } b = 9\).

(i) First, find \(\overrightarrow{AB}\) by subtracting \(\overrightarrow{OA}\) from \(\overrightarrow{OB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix} - \begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -6 \end{pmatrix}\).

Then, find \(\overrightarrow{AP}\):

\(\overrightarrow{AP} = \frac{1}{3} \overrightarrow{AB} = \frac{1}{3} \begin{pmatrix} 0 \\ 3 \\ -6 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}\).

Now, find \(\overrightarrow{OP}\) by adding \(\overrightarrow{OA}\) and \(\overrightarrow{AP}\):

\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix}\).

(ii) The distance \(OP\) is the magnitude of \(\overrightarrow{OP}\):

\(OP = \sqrt{5^2 + 2^2 + 1^2} = \sqrt{30} \approx 5.48\).

(iii) To check if \(OP\) is perpendicular to \(AB\), calculate the dot product \(\overrightarrow{AB} \cdot \overrightarrow{OP}\):

\(\overrightarrow{AB} \cdot \overrightarrow{OP} = \begin{pmatrix} 0 \\ 3 \\ -6 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix} = 0 \times 5 + 3 \times 2 + (-6) \times 1 = 0 + 6 - 6 = 0\).

Since the dot product is zero, \(OP\) is perpendicular to \(AB\).

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

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos AOB\)

Given \(p = 2\), \(\overrightarrow{OA} = 3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\) and \(\overrightarrow{OB} = 6\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}\).

Calculate the dot product:

\(3 \times 6 + 2 \times 6 + (-4) \times 3 = 18\)

Calculate magnitudes:

\(|\overrightarrow{OA}| = \sqrt{3^2 + 2^2 + (-4)^2} = \sqrt{29}\)

\(|\overrightarrow{OB}| = \sqrt{6^2 + 6^2 + 3^2} = 9\)

Substitute into the formula:

\(\sqrt{29} \times 9 \times \cos AOB = 18\)

\(\cos AOB = \frac{18}{\sqrt{29} \times 9}\)

\(AOB = \cos^{-1}\left(\frac{18}{\sqrt{29} \times 9}\right) \approx 68.2^\circ\) or \(1.19 \text{ radians}\)

(ii) For \(\overrightarrow{AB}\) to be parallel to \(\overrightarrow{OC}\), \(\overrightarrow{AB} = \lambda \overrightarrow{OC}\) for some scalar \(\lambda\).

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (6 - 3)\mathbf{i} + ((p + 4) - p)\mathbf{j} + (3 + 2p)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} + (3 + 2p)\mathbf{k}\)

\(\overrightarrow{OC} = (p - 1)\mathbf{i} + 2\mathbf{j} + q\mathbf{k}\)

Comparing the \(\mathbf{j}\) components:

\(4 = 2 \lambda \Rightarrow \lambda = 2\)

Comparing the \(\mathbf{i}\) components:

\(3 = 2(p - 1) \Rightarrow 3 = 2p - 2 \Rightarrow 2p = 5 \Rightarrow p = 2.5\)

Comparing the \(\mathbf{k}\) components:

\(3 + 2p = 2q \Rightarrow 3 + 2(2.5) = 2q \Rightarrow 8 = 2q \Rightarrow q = 4\)

Thus, \(p = 2.5\) and \(q = 4\).

(i) Since \(\angle AOB = 90^\circ\), the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

\(\overrightarrow{OA} = \begin{pmatrix} 3 \\ -6 \\ p \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 2 \\ -6 \\ -7 \end{pmatrix}\).

Calculate the dot product:

\(3 \times 2 + (-6) \times (-6) + p \times (-7) = 0\)

\(6 + 36 - 7p = 0\)

\(42 = 7p\)

\(p = 6\)

(ii) \(\overrightarrow{OC} = \frac{2}{3} \overrightarrow{OA} = \frac{2}{3} \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ -6 \\ -7 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 11 \end{pmatrix}\)

Magnitude of \(\overrightarrow{BC} = \sqrt{0^2 + 2^2 + 11^2} = \sqrt{125}\)

Unit vector in the direction of \(\overrightarrow{BC} = \frac{1}{\sqrt{125}} \begin{pmatrix} 0 \\ 2 \\ 11 \end{pmatrix}\)