(a) The position vector of M is the midpoint of EF. Since E is at (3, 2, 0) and F is at (3, 2, 2), the midpoint M is at (3, 2, 1). Therefore, the position vector of M is \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\).

(b) To calculate angle PAM, find vectors \(\overrightarrow{AM}\) and \(\overrightarrow{AP}\):

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

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

Calculate the dot product: \(\overrightarrow{AM} \cdot \overrightarrow{AP} = 0 + 2 \times 1 + 1 \times 2 = 4\)

Find the magnitudes: \(|\overrightarrow{AM}| = \sqrt{2^2 + 1^2} = \sqrt{5}\), \(|\overrightarrow{AP}| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{9} = 3\)

Use the cosine formula: \(\cos \theta = \frac{4}{\sqrt{5} \times 3}\)

\(\theta = \cos^{-1}\left(\frac{4}{3\sqrt{5}}\right) \approx 53.4^\circ\)

(c) To find the perpendicular from P to the line through O and M, express PQ for a general point Q on the line:

\(\overrightarrow{PQ} = (\mathbf{i} + \mathbf{j} + 2\mathbf{k}) + \mu(3\mathbf{i} + 2\mathbf{j} + \mathbf{k})\)

Calculate the scalar product of \(\overrightarrow{PQ}\) and the direction vector \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and equate to zero to find \(\mu\).

Solve to find \(\mu = -\frac{1}{2}\).

Calculate \(PQ\) using \(\mu\):

\(PQ = \sqrt{5 + 0 + 1.5^2} = \frac{\sqrt{10}}{2}\)

(i) To find \(\overrightarrow{OE}\), note that E divides OB in the ratio \(\frac{2}{10}\) since \(OB = \sqrt{8^2 + 6^2} = 10\). Thus, \(\overrightarrow{OE} = \frac{2}{10}(8\mathbf{i} + 6\mathbf{j}) = 1.6\mathbf{i} + 1.2\mathbf{j}\).

(ii) First, find \(\overrightarrow{OD}\) and \(\overrightarrow{BD}\):

\(\overrightarrow{OD} = 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k}\)

\(\overrightarrow{BD} = -8\mathbf{i} - 6\mathbf{j} + 1.6\mathbf{i} + 1.2\mathbf{j} + 7\mathbf{k} = -6.4\mathbf{i} - 4.8\mathbf{j} + 7\mathbf{k}\)

Calculate the magnitudes:

\(|\overrightarrow{OD}| = \sqrt{1.6^2 + 1.2^2 + 7^2} = \sqrt{53}\)

\(|\overrightarrow{BD}| = \sqrt{(-6.4)^2 + (-4.8)^2 + 7^2} = \sqrt{113}\)

Find the dot product:

\(\overrightarrow{OD} \cdot \overrightarrow{BD} = 1.6(-6.4) + 1.2(-4.8) + 7(7) = -10.24 - 5.76 + 49 = 33\)

Use the scalar product to find the angle:

\(\cos \angle BDO = \frac{\overrightarrow{OD} \cdot \overrightarrow{BD}}{|\overrightarrow{OD}| \times |\overrightarrow{BD}|} = \frac{33}{\sqrt{53} \times \sqrt{113}}\)

\(\angle BDO = \cos^{-1}\left(\frac{33}{77.4}\right) \approx 64.8^{\circ}\)

(i) To find \(\overrightarrow{DA}\), we calculate:

\(\overrightarrow{DA} = \overrightarrow{OA} - \overrightarrow{OD} = (8\mathbf{i}) - (2\mathbf{i} + 4\mathbf{k}) = 6\mathbf{i} - 4\mathbf{k}\).

To find \(\overrightarrow{CA}\), we calculate:

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

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

\(\overrightarrow{CA} \cdot \overrightarrow{DA} = (6\mathbf{i} - 5\mathbf{j} - 4\mathbf{k}) \cdot (6\mathbf{i} - 4\mathbf{k}) = 36 + 16 = 52\).

Calculate the magnitudes:

\(|\overrightarrow{DA}| = \sqrt{6^2 + 0^2 + (-4)^2} = \sqrt{52}\).

\(|\overrightarrow{CA}| = \sqrt{6^2 + (-5)^2 + (-4)^2} = \sqrt{77}\).

Using the cosine formula:

\(52 = \sqrt{77} \cdot \sqrt{52} \cdot \cos \angle CAD\).

\(\cos \angle CAD = \frac{52}{\sqrt{77} \cdot \sqrt{52}} \approx 0.82178\).

\(\angle CAD \approx 34.7^\circ\) or \(0.606 \text{ radians}\).

(i) To find \(\overrightarrow{CE}\), we use the coordinates of C and E. Since E is 8 units above D, and D is at (0, 2, 0), E is at (0, 2, 8). C is at (4, 1, 0). Thus, \(\overrightarrow{CE} = (0 - 4)\mathbf{i} + (2 - 1)\mathbf{j} + (8 - 0)\mathbf{k} = -4\mathbf{i} - \mathbf{j} + 8\mathbf{k}\).

The length of \(CE\) is \(|CE| = \sqrt{(-4)^2 + (-1)^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9\).

(ii) To find angle ECA, we first find \(\overrightarrow{CA}\). A is at (7, 0, 0), so \(\overrightarrow{CA} = (7 - 4)\mathbf{i} + (0 - 1)\mathbf{j} = 3\mathbf{i} - \mathbf{j}\).

The scalar product \(\overrightarrow{CE} \cdot \overrightarrow{CA} = (-4\mathbf{i} - \mathbf{j} + 8\mathbf{k}) \cdot (3\mathbf{i} - \mathbf{j}) = -12 + 1 = -11\).

The magnitudes are \(|\overrightarrow{CE}| = 9\) and \(|\overrightarrow{CA}| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}\).

Thus, \(\cos \theta = \frac{-11}{9\sqrt{10}}\), and \(\theta = \cos^{-1} \left( \frac{-11}{9\sqrt{10}} \right)\).

Rewriting in the required form, \(\cos^{-1} \left( \frac{-1}{\sqrt{18}} \right)\).

(i) To show \(OAB = 90^\circ\), calculate \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 8 \\ 3 \\ 2 \end{pmatrix}\).

Calculate the dot product \(\overrightarrow{OA} \cdot \overrightarrow{AB} = \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 3 \\ 2 \end{pmatrix} = -16 - 6 - 2 = -24\).

Since the dot product is zero, \(OAB = 90^\circ\).

(ii) The magnitude of \(\overrightarrow{OA}\) is \(\sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3\).

The magnitude of \(\overrightarrow{CB}\) is \(3 \times 3 = 9\).

\(\overrightarrow{CB} = \begin{pmatrix} 6 \\ -6 \\ -3 \end{pmatrix}\) or \(\overrightarrow{BC} = \begin{pmatrix} -6 \\ 6 \\ 3 \end{pmatrix}\).

\(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} = \begin{pmatrix} 6 \\ 1 \\ 1 \end{pmatrix} + \begin{pmatrix} -6 \\ 6 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \\ 4 \end{pmatrix}\).

(iii) The area of trapezium \(OABC\) is \(\frac{1}{2} \times (3 + 9) \times \sqrt{29} = 6\sqrt{29}\).