(i) To find \(\overrightarrow{PB}\), we first find \(\overrightarrow{P}\) as the midpoint of \(\overrightarrow{OD}\):

\(\overrightarrow{P} = \frac{1}{2}(\overrightarrow{O} + \overrightarrow{D}) = \frac{1}{2}(2\mathbf{i} + 10\mathbf{k}) = \mathbf{i} + 5\mathbf{k}\)

\(\overrightarrow{B} = \overrightarrow{OA} + \overrightarrow{OC} = 6\mathbf{i} + 8\mathbf{j}\)

\(\overrightarrow{PB} = \overrightarrow{B} - \overrightarrow{P} = (6\mathbf{i} + 8\mathbf{j}) - (\mathbf{i} + 5\mathbf{k}) = 5\mathbf{i} + 8\mathbf{j} - 5\mathbf{k}\)

To find \(\overrightarrow{PQ}\), we first find \(\overrightarrow{Q}\) as the midpoint of \(\overrightarrow{GF}\):

\(\overrightarrow{Q} = \frac{1}{2}(\overrightarrow{G} + \overrightarrow{F}) = \frac{1}{2}(6\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)

\(\overrightarrow{PQ} = \overrightarrow{Q} - \overrightarrow{P} = (3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - (\mathbf{i} + 5\mathbf{k}) = 4\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}\)

(ii) Calculate the lengths:

\(\text{Length of } PB = \sqrt{5^2 + 8^2 + (-5)^2} = \sqrt{114} \approx 10.7\)

\(\text{Length of } PQ = \sqrt{4^2 + 8^2 + 5^2} = \sqrt{105} \approx 10.2\)

Since \(10.2 < 10.7\), P is nearer to Q.

(iii) Use the scalar product to find angle \(BPQ\):

\(\overrightarrow{PB} \cdot \overrightarrow{PQ} = 5 \times 4 + 8 \times 8 + (-5) \times 5 = 20 + 64 - 25 = 59\)

\(\text{Magnitude of } \overrightarrow{PB} = \sqrt{114}\), \(\text{Magnitude of } \overrightarrow{PQ} = \sqrt{105}\)

\(\cos BPQ = \frac{59}{\sqrt{114} \times \sqrt{105}}\)

\(BPQ = \cos^{-1}\left(\frac{59}{\sqrt{114} \times \sqrt{105}}\right) \approx 57.4^\circ \text{ or } 1.00 \text{ rad}\)