(i) To find \(\overrightarrow{OB}\), use \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) + (3\mathbf{i} - \mathbf{j} + \mathbf{k}) = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).

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

The unit vector in the direction of \(\overrightarrow{OB}\) is \(\frac{1}{6}(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).

(ii) To find \(\overrightarrow{AC}\), use \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3\mathbf{i} - \mathbf{j} + \mathbf{k}) - (\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}\).

The magnitudes are \(|\overrightarrow{OB}| = 6\) and \(|\overrightarrow{AC}| = \sqrt{24}\).

The dot product \(\overrightarrow{AC} \cdot \overrightarrow{OB} = 8 - 8 - 8 = -8\).

Using \(\overrightarrow{AC} \cdot \overrightarrow{OB} = |\overrightarrow{AC}| |\overrightarrow{OB}| \cos \theta\), we have \(-8 = 6 \times \sqrt{24} \times \cos \theta\).

Solving for \(\theta\), \(\theta = 105.8^\circ\), so the acute angle is \(74.2^\circ\).

(iii) The magnitudes are \(|\overrightarrow{OA}| = \sqrt{19}\) and \(|\overrightarrow{OC}| = \sqrt{11}\).

The perimeter is \(2(\sqrt{19} + \sqrt{11}) \approx 15.4\).

(i) To find \(\overrightarrow{OQ}\), note that \(Q\) is the midpoint of \(DF\). Since \(D = (0, 6, 6)\) and \(F = (6, 6, 6)\), the midpoint \(Q\) is \((3, 6, 6)\). Therefore, \(\overrightarrow{OQ} = 3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\).

For \(\overrightarrow{PQ}\), since \(P\) is such that \(\overrightarrow{AP} = \frac{1}{3} \overrightarrow{AB}\), and \(A = (6, 0, 0)\), \(B = (6, 6, 0)\), we have \(P = (6, 2, 0)\). Thus, \(\overrightarrow{PQ} = (3 - 6)\mathbf{i} + (6 - 2)\mathbf{j} + (6 - 0)\mathbf{k} = -3\mathbf{i} + \mathbf{j} + 6\mathbf{k}\).

(ii) To find the angle \(OQP\), use the dot product formula: \(\overrightarrow{OQ} \cdot \overrightarrow{PQ} = |\overrightarrow{OQ}| |\overrightarrow{PQ}| \cos \theta\).

Calculate \(\overrightarrow{OQ} \cdot \overrightarrow{PQ} = (3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) \cdot (-3\mathbf{i} + \mathbf{j} + 6\mathbf{k}) = -9 + 3 + 36 = 30\).

Find magnitudes: \(|\overrightarrow{OQ}| = \sqrt{3^2 + 3^2 + 6^2} = \sqrt{54}\), \(|\overrightarrow{PQ}| = \sqrt{(-3)^2 + 1^2 + 6^2} = \sqrt{46}\).

Thus, \(30 = \sqrt{54} \sqrt{46} \cos \theta\).

\(\cos \theta = \frac{30}{\sqrt{54} \sqrt{46}}\).

\(\theta = \cos^{-1}\left(\frac{30}{\sqrt{54} \sqrt{46}}\right) \approx 53.0^\circ\).

(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 \(BPQ\):

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

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

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

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

(i) To find \(\overrightarrow{PA}\), note that \(P\) is 8 cm along \(EF\) from \(E\), so \(P\) has coordinates \((8, 0, 6)\). \(A\) is at \((0, 0, 0)\). Thus, \(\overrightarrow{PA} = (0 - 8)\mathbf{i} + (0 - 0)\mathbf{j} + (0 - 6)\mathbf{k} = -8\mathbf{i} - 6\mathbf{k}\).

For \(\overrightarrow{PN}\), \(N\) is at \((20, 0, 0)\). Thus, \(\overrightarrow{PN} = (20 - 8)\mathbf{i} + (0 - 0)\mathbf{j} + (0 - 6)\mathbf{k} = 12\mathbf{i} - 6\mathbf{k}\).

(ii) The dot product \(\overrightarrow{PA} \cdot \overrightarrow{PN} = (-8)(12) + (0)(0) + (-6)(-6) = -96 + 36 = -60\).

The magnitudes are \(|\overrightarrow{PA}| = \sqrt{(-8)^2 + 0^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\).

\(|\overrightarrow{PN}| = \sqrt{(12)^2 + 0^2 + (-6)^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5}\).

\(\cos \angle APN = \frac{-60}{10 \times 6\sqrt{5}} = \frac{-60}{60\sqrt{5}} = -\frac{1}{\sqrt{5}}\).

\(\angle APN = \cos^{-1}\left(-\frac{1}{\sqrt{5}}\right) \approx 99^\circ\).

(i) The coordinates of P are \((2, 0, 0)\) since it is the midpoint of OA. The coordinates of Q are \((0, 2, 4)\) since it is the midpoint of DG. The coordinates of R are \((2, 2, 2)\) since it is the center of the face ABFE.

\(\overrightarrow{PR} = (2 - 2)\mathbf{i} + (2 - 0)\mathbf{j} + (2 - 0)\mathbf{k} = 2\mathbf{j} + 2\mathbf{k}\)

\(\overrightarrow{PQ} = (0 - 2)\mathbf{i} + (2 - 0)\mathbf{j} + (4 - 0)\mathbf{k} = -2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\)

(ii) The dot product \(\overrightarrow{PQ} \cdot \overrightarrow{PR} = (-2)(0) + (2)(2) + (4)(2) = 8\)

The magnitudes are \(|\overrightarrow{PQ}| = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{24}\) and \(|\overrightarrow{PR}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8}\).

Using the dot product formula, \(\cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PR}}{|\overrightarrow{PQ}| |\overrightarrow{PR}|} = \frac{8}{\sqrt{24} \cdot \sqrt{8}}\)

\(\theta = \cos^{-1}\left(\frac{8}{\sqrt{24} \cdot \sqrt{8}}\right) \approx 61.9^\circ\)

(iii) \(\overrightarrow{QR} = (2 - 0)\mathbf{i} + (2 - 2)\mathbf{j} + (2 - 4)\mathbf{k} = 2\mathbf{i} - 2\mathbf{k}\)

\(|\overrightarrow{QR}| = \sqrt{2^2 + 0^2 + (-2)^2} = \sqrt{8}\)

Perimeter of \(\triangle PQR = |\overrightarrow{PR}| + |\overrightarrow{PQ}| + |\overrightarrow{QR}| = \sqrt{8} + \sqrt{24} + \sqrt{8} \approx 12.8 \text{ cm}\)