(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 \(\overrightarrow{BC} = \frac{1}{\sqrt{125}} \begin{pmatrix} 0 \\ 2 \\ 11 \end{pmatrix}\)

(i) First, find the vector \(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = -5\mathbf{i} - \mathbf{j} + 2\mathbf{k}\).

Calculate the dot product \(\overrightarrow{OA} \cdot \overrightarrow{BA} = (2)(-5) + (3)(-1) + (5)(2) = -10 - 3 + 10 = -3\).

Find the magnitudes: \(|\overrightarrow{OA}| = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{38}\) and \(|\overrightarrow{BA}| = \sqrt{(-5)^2 + (-1)^2 + 2^2} = \sqrt{30}\).

Use the cosine formula: \(\cos OAB = \frac{-3}{\sqrt{38} \times \sqrt{30}}\).

Calculate \(OAB = \cos^{-1}\left(\frac{-3}{\sqrt{38} \times \sqrt{30}}\right) \approx 95.1^\circ\) (or 1.66 radians).

(ii) The area of triangle \(OAB\) is given by \(\frac{1}{2} |\overrightarrow{OA}| |\overrightarrow{BA}| \sin OAB\).

Substitute the values: \(\frac{1}{2} \times \sqrt{38} \times \sqrt{30} \times \sin 95.1^\circ \approx 16.8\).

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

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

Calculate the dot product: \(2(-2) + (-2)(3) + (-1)(6) = -16\)

Calculate magnitudes: \(|\overrightarrow{OA}| = \sqrt{2^2 + (-2)^2 + (-1)^2} = 3\)

\(|\overrightarrow{OB}| = \sqrt{(-2)^2 + 3^2 + 6^2} = 7\)

Substitute into the formula: \(3 \times 7 \times \cos \theta = -16\)

Solve for \(\theta\): \(\theta = 139.6^\circ\) or \(2.44^c\) or \(0.776\pi\)

(ii) Find \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ 6 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix}\)

Magnitude of \(\overrightarrow{AC} = \sqrt{0^2 + 8^2 + 6^2} = 10\)

Scale to magnitude 15: \(\frac{15}{10} \times \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 9 \end{pmatrix}\)

(iii) For \(p\overrightarrow{OA} + \overrightarrow{OC}\) to be perpendicular to \(\overrightarrow{OB}\), the dot product must be zero:

\((p\overrightarrow{OA} + \overrightarrow{OC}) \cdot \overrightarrow{OB} = 0\)

\(\begin{pmatrix} 2+2p \\ 6-2p \\ 5-p \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 3 \\ 6 \end{pmatrix} = 0\)

\(-2(2+2p) + 3(6-2p) + 6(5-p) = 0\)

Simplify: \(-4 - 4p + 18 - 6p + 30 - 6p = 0\)

\(44 - 16p = 0\)

\(p = \frac{2}{3/4}\)

(i) To find the value of \(p\) for which the lengths of \(AB\) and \(CB\) are equal, we first find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 1 \\ 5 \\ p \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ p+4 \end{pmatrix}\)

\(\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = \begin{pmatrix} 1 \\ 5 \\ p \end{pmatrix} - \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 \\ 5 \\ p-2 \end{pmatrix}\)

Equating the magnitudes:

\(\sqrt{(-1)^2 + 2^2 + (p+4)^2} = \sqrt{(-4)^2 + 5^2 + (p-2)^2}\)

\(1 + 4 + (p+4)^2 = 16 + 25 + (p-2)^2\)

\(1 + 4 + (p+4)^2 = 16 + 25 + (p-2)^2\)

\((p+4)^2 = 41 + (p-2)^2\)

Solving gives \(p = 2\).

(ii) For \(p = 1\), find \(\angle ABC\) using the scalar product:

\(\overrightarrow{AB} = \begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix}\)

\(\overrightarrow{CB} = \begin{pmatrix} -4 \\ 5 \\ -1 \end{pmatrix}\)

\(\overrightarrow{AB} \cdot \overrightarrow{CB} = (-1)(-4) + (2)(5) + (5)(-1) = 4 + 10 - 5 = 9\)

\(|\overrightarrow{AB}| = \sqrt{1 + 4 + 25} = \sqrt{30}\)

\(|\overrightarrow{CB}| = \sqrt{16 + 25 + 1} = \sqrt{42}\)

\(\cos \angle ABC = \frac{9}{\sqrt{30} \times \sqrt{42}}\)

\(\angle ABC = \cos^{-1}\left(\frac{9}{\sqrt{30} \times \sqrt{42}}\right) \approx 75.3^\circ\) or \(1.31 \text{ radians}\).

Given \(\overrightarrow{OA} = 2\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}\) and \(\overrightarrow{OB} = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\).

First, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4 - 2)\mathbf{i} + (-4 + 5)\mathbf{j} + (2 + 2)\mathbf{k} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}\).

Since \(\overrightarrow{AB} = \overrightarrow{BC}\), we have \(\overrightarrow{AC} = 2\overrightarrow{AB} = 4\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}\).

Thus, \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = (2\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}) + (4\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}) = 6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\).

The magnitude of \(\overrightarrow{OC}\) is \(\sqrt{6^2 + (-3)^2 + 6^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9\).

Therefore, the unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{9}(6\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})\).

(i) To find \(k\) when \(\angle AOB = 90^\circ\), use the scalar product: \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

\(\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -1 \\ k \end{pmatrix} = 2 \times 5 + 1 \times (-1) + (-2) \times k = 0\)

\(10 - 1 - 2k = 0\)

\(9 = 2k\)

\(k = \frac{9}{2}\)

(ii) For \(|\overrightarrow{AB}| = |\overrightarrow{OC}|\), first find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ -1 \\ k \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ k+2 \end{pmatrix}\).

\(|\overrightarrow{OC}| = \sqrt{2^2 + 6^2 + (-3)^2} = 7\).

Equate magnitudes: \(\sqrt{3^2 + (-2)^2 + (k+2)^2} = 7\).

\(9 + 4 + (k+2)^2 = 49\)

\((k+2)^2 = 36\)

\(k+2 = 6\) or \(k+2 = -6\)

\(k = 4\) or \(k = -8\)

(iii) \(\overrightarrow{OD} = 3 \overrightarrow{OA} = 3 \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ -6 \end{pmatrix}\).

\(\overrightarrow{OE} = 2 \overrightarrow{OC} = 2 \begin{pmatrix} 2 \\ 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 12 \\ -6 \end{pmatrix}\).

\(\overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} = \begin{pmatrix} 4 \\ 12 \\ -6 \end{pmatrix} - \begin{pmatrix} 6 \\ 3 \\ -6 \end{pmatrix} = \begin{pmatrix} -2 \\ 9 \\ 0 \end{pmatrix}\).

Magnitude of \(\overrightarrow{DE} = \sqrt{(-2)^2 + 9^2 + 0^2} = \sqrt{4 + 81} = \sqrt{85}\).

(i) To find the value of \(p\) when \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OB}\), set the scalar product to zero:

\((p-6)(4-2p) + (2p-6)p + 1 \times 2 = 0\)

\(-2p^2 + 16p - 24 + 2p^2 - 6p + 2 = 0\)

\(10p - 22 = 0\)

Solving for \(p\), we get \(p = 2.2\).

(ii) For \(OAB\) to be a straight line, \(\overrightarrow{OB} = 2 \overrightarrow{OA}\):

\(4-2p = 2(p-6)\) and \(p = 2(2p-6)\)

Solving \(4-2p = 2(p-6)\), we get \(p = 4\).

Thus, \(\overrightarrow{OA} = \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} -4 \\ 4 \\ 2 \end{pmatrix}\).

The length of \(OA\) is \(\sqrt{(-2)^2 + 2^2 + 1^2} = 3\).

(a) To find the cosine of angle BAC, we first find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ 3 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix}\)

The scalar product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 2 \times 4 + (-2) \times 0 + 1 \times (-3) = 8 - 3 = 5\).

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\) and \(|\overrightarrow{AC}| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{25} = 5\).

Thus, \(\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{5}{3 \times 5} = \frac{1}{3}\).

(b) To find the area of triangle ABC, we use the formula \(\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|\).

The cross product \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ 4 & 0 & -3 \end{vmatrix} = \mathbf{i}((-2)(-3) - 1(0)) - \mathbf{j}(2(-3) - 1(4)) + \mathbf{k}(2(0) - (-2)(4))\).

\(= \mathbf{i}(6) - \mathbf{j}(-6 - 4) + \mathbf{k}(8) = 6\mathbf{i} + 10\mathbf{j} + 8\mathbf{k}\).

The magnitude \(|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{6^2 + 10^2 + 8^2} = \sqrt{36 + 100 + 64} = \sqrt{200} = 10\sqrt{2}\).

Thus, the area is \(\frac{1}{2} \times 10\sqrt{2} = 5\sqrt{2}\).

(i) For \(ABC\) to be a straight line, vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) must be parallel. Calculate \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} 1 \\ p-5 \\ q+2 \end{pmatrix}\). Equating the ratios gives:

\(\frac{2}{1} = \frac{3}{p-5} = \frac{1}{q+2}\)

Solving these gives \(p = 6.5\) and \(q = -1.5\).

(ii) For \(\angle BAC = 90^\circ\), vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) must be perpendicular. Calculate \(\overrightarrow{AC} = \begin{pmatrix} 3 \\ p-2 \\ q+3 \end{pmatrix}\). The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 0\) gives:

\(6 + 3(p-2) + (q+3) = 0\)

Solving gives \(q = -3p - 3\).

(iii) For \(p = 3\) and \(AB = AC\), equate the magnitudes:

\(AB^2 = 4 + 9 + 1 = 14\)

\(AC^2 = 9 + 1 + (q+3)^2\)

Equating gives \((q+3)^2 = 4\), so \(q+3 = \pm 2\).

Thus, \(q = -1\) or \(q = -5\).

(i) To show that \(\angle ABC = 90^\circ\), we need to show that vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are perpendicular.

First, find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} - \begin{pmatrix} 3 \\ 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}.\)

Next, find \(\overrightarrow{BC}\):

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

Calculate the dot product \(\overrightarrow{AB} \cdot \overrightarrow{BC}\):

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 2 \times 1 + (-3) \times 2 + 1 \times 4 = 2 - 6 + 4 = 0.\)

Since the dot product is zero, \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are perpendicular, so \(\angle ABC = 90^\circ\).

(ii) To find the area of triangle \(ABC\), use the formula:

\(\text{Area} = \frac{1}{2} \times |\overrightarrow{AB}| \times |\overrightarrow{BC}|.\)

Calculate the magnitudes:

\(|\overrightarrow{AB}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}.\)

\(|\overrightarrow{BC}| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{21}.\)

Thus, the area is:

\(\text{Area} = \frac{1}{2} \times \sqrt{14} \times \sqrt{21} = \frac{1}{2} \times \sqrt{294}.\)

\(\sqrt{294} \approx 17.146,\) so the area is approximately \(8.6\) to 1 decimal place.

(i) Calculate the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 6 + 4 + 16 = 26\).

Find the magnitudes: \(|\overrightarrow{OA}| = \sqrt{36} = 6\) and \(|\overrightarrow{OB}| = \sqrt{26}\).

Use the cosine formula: \(\cos AOB = \frac{26}{6\sqrt{26}}\).

Calculate \(AOB\): \(AOB = \cos^{-1}\left(\frac{26}{6\sqrt{26}}\right) \approx 31.8^{\circ}\).

(ii) Find \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}\).

Calculate \(\overrightarrow{OC} = \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix} + 2\begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}\).

Find the unit vector: \(\frac{1}{6} \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}\).

(iii) Calculate \(|\overrightarrow{OC}| = \sqrt{16 + 4 + 16} = 6\).

Since \(|\overrightarrow{OC}| = |\overrightarrow{OA}| = 6\), triangle OAC is isosceles.

(i) To find the cosine of angle \(AOB\), we use the dot product formula:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 3 \times 6 + 0 \times (-3) + (-4) \times 2 = 18 - 8 = 10\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{3^2 + 0^2 + (-4)^2} = 5\) and \(|\overrightarrow{OB}| = \sqrt{6^2 + (-3)^2 + 2^2} = 7\).

Thus, \(\cos \angle AOB = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}| |\overrightarrow{OB}|} = \frac{10}{35} = \frac{2}{7}\).

(ii) The vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ 6 \end{pmatrix}\).

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

Given \(|\overrightarrow{AB}| = |\overrightarrow{OC}|\), we have:

\(\sqrt{k^2 + (-2k)^2 + (2k - 3)^2} = 7\).

Squaring both sides, \(k^2 + 4k^2 + (2k - 3)^2 = 49\).

\(k^2 + 4k^2 + 4k^2 - 12k + 9 = 49\).

\(9k^2 - 12k - 40 = 0\).

Solving the quadratic equation, \(k = 3\) or \(k = -\frac{5}{3}\).

(i) To find when \(\overrightarrow{OA}\) is perpendicular to \(\overrightarrow{OB}\), their dot product must be zero:

\((1)(-7) + (3)(1-p) + (p)(p) = 0\)

\(-7 + 3 - 3p + p^2 = 0\)

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

Factoring gives \((p+1)(p-4) = 0\), so \(p = -1\) or \(p = 4\).

(ii) The magnitudes are \(a = \sqrt{1^2 + 3^2 + p^2}\) and \(b = \sqrt{(-7)^2 + (1-p)^2 + p^2}\).

Given \(b^2 = 2a^2\):

\(49 + (1-p)^2 + p^2 = 2(1 + 9 + p^2)\)

\(49 + 1 - 2p + p^2 + p^2 = 2(10 + p^2)\)

\(50 - 2p + 2p^2 = 20 + 2p^2\)

\(50 - 2p = 20\)

\(2p = 30\)

\(p = 15\).

(iii) When \(p = -8\), \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (-7\mathbf{i} + 9\mathbf{j} + (-8)\mathbf{k}) - (\mathbf{i} + 3\mathbf{j} - 8\mathbf{k})\).

\(\overrightarrow{AB} = -8\mathbf{i} + 6\mathbf{j}\).

The magnitude \(|\overrightarrow{AB}| = \sqrt{(-8)^2 + 6^2} = 10\).

The unit vector is \(\frac{1}{10}(-8\mathbf{i} + 6\mathbf{j})\).

(i) To show that angle OAB is a right angle, we need to show that vectors OA and AB are perpendicular. The position vector of A is 3i + 2j - k and the position vector of B is 7i - 3j + k.

The vector AB is given by:

\(\mathbf{AB} = \mathbf{B} - \mathbf{A} = (7\mathbf{i} - 3\mathbf{j} + \mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 4\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}\)

The dot product \(\mathbf{AO} \cdot \mathbf{AB}\) is:

\((3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \cdot (4\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}) = 3 \times 4 + 2 \times (-5) + (-1) \times 2 = 12 - 10 - 2 = 0\)

Since the dot product is zero, angle OAB is a right angle.

(ii) To find the area of triangle OAB, we use the formula for the area of a triangle given by vectors:

\(\text{Area} = \frac{1}{2} \times |\mathbf{OA}| \times |\mathbf{AB}|\)

First, calculate the magnitudes:

\(|\mathbf{OA}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\)

\(|\mathbf{AB}| = \sqrt{4^2 + (-5)^2 + 2^2} = \sqrt{16 + 25 + 4} = \sqrt{45}\)

Thus, the area is:

\(\text{Area} = \frac{1}{2} \times \sqrt{14} \times \sqrt{45} = \frac{1}{2} \times \sqrt{630} = 12.5\)

(i) To find angle \(BAC\), we need vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ -1 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}\)

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = 4 \times 0 + (-2) \times 3 + 4 \times 4 = 0 - 6 + 16 = 10\)

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{36} = 6\)

\(|\overrightarrow{AC}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{25} = 5\)

Using the cosine formula: \(\cos BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| \times |\overrightarrow{AC}|} = \frac{10}{6 \times 5} = \frac{1}{3}\)

Thus, \(BAC = \cos^{-1}\left(\frac{1}{3}\right)\).

(ii) To find the area of triangle \(ABC\), use \(\sin BAC = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}\)

Area = \(\frac{1}{2} \times |\overrightarrow{AB}| \times |\overrightarrow{AC}| \times \sin BAC = \frac{1}{2} \times 6 \times 5 \times \frac{\sqrt{8}}{3} = 5\sqrt{8}\)

(i) To find the values of \(p\) for which angle \(AOB\) is 90°, we need \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

Calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3p)(-p) + (4)(-1) + (p^2)(p^2) = -3p^2 - 4 + p^4\).

Set the dot product to zero:

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

Rearrange to form a quadratic in \(p^2\):

\(p^4 - 3p^2 - 4 = 0\).

Factorize:

\((p^2 + 1)(p^2 - 4) = 0\).

Thus, \(p^2 = -1\) (no real solution) or \(p^2 = 4\).

Therefore, \(p = \pm 2\).

(ii) For \(p = 3\), find \(\overrightarrow{BA}\):

\(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} 3(3) \\ 4 \\ 3^2 \end{pmatrix} - \begin{pmatrix} -3 \\ -1 \\ 9 \end{pmatrix} = \begin{pmatrix} 9 \\ 4 \\ 9 \end{pmatrix} - \begin{pmatrix} -3 \\ -1 \\ 9 \end{pmatrix} = \begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}\).

Calculate the magnitude of \(\overrightarrow{BA}\):

\(|\overrightarrow{BA}| = \sqrt{12^2 + 5^2 + 0^2} = \sqrt{144 + 25} = \sqrt{169} = 13\).

The unit vector in the direction of \(\overrightarrow{BA}\) is:

\(\frac{1}{13} \begin{pmatrix} 12 \\ 5 \\ 0 \end{pmatrix}\).

(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, \(\overrightarrow{OC} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\).

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

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 1 \cdot k + 0 \cdot (-k) + 2 \cdot 2k = k + 4k = 5k\).

For \(k = 2\), \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 10\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{5}\) and \(|\overrightarrow{OB}| = \sqrt{k^2 + (-k)^2 + (2k)^2} = \sqrt{6k^2} = \sqrt{24}\).

Using the dot product formula, \(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| \cdot |\overrightarrow{OB}| \cdot \cos \theta\).

\(10 = \sqrt{5} \cdot \sqrt{24} \cdot \cos \theta\).

\(\cos \theta = \frac{10}{\sqrt{5} \cdot \sqrt{24}}\).

\(\theta = \cos^{-1}\left(\frac{10}{\sqrt{5} \cdot \sqrt{24}}\right) \approx 24.1^\circ\).

(ii) For \(\overrightarrow{AB}\) to be a unit vector, its magnitude must be 1.

\(\overrightarrow{AB} = \begin{pmatrix} k-1 \\ -k \\ 2k-2 \end{pmatrix}\).

Magnitude: \(\sqrt{(k-1)^2 + (-k)^2 + (2k-2)^2} = 1\).

\((k-1)^2 + k^2 + (2k-2)^2 = 1\).

\((k-1)^2 = k^2 - 2k + 1\), \((2k-2)^2 = 4k^2 - 8k + 4\).

\(k^2 - 2k + 1 + k^2 + 4k^2 - 8k + 4 = 1\).

\(6k^2 - 10k + 5 = 1\).

\(6k^2 - 10k + 4 = 0\).

Solving the quadratic equation: \(k = 1\) or \(k = \frac{2}{3}\).

(i) To find \(\overrightarrow{CD}\), calculate \(3\mathbf{a} - 2\mathbf{b}\):

\(3\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} - 2\begin{pmatrix} 4 \\ 0 \\ 6 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} - \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

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

Thus, the unit vector is \(\frac{1}{7} \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}\).

(ii) The position vector of E is the midpoint of C and D:

\(\overrightarrow{OE} = \frac{1}{2}(3\mathbf{a} + 2\mathbf{b}) = \frac{1}{2}\left(\begin{pmatrix} 6 \\ 3 \\ 6 \end{pmatrix} + \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix}\right) = \begin{pmatrix} 7 \\ 1.5 \\ 9 \end{pmatrix}\).

Calculate \(\overrightarrow{OE} \cdot \overrightarrow{OD}\):

\(\begin{pmatrix} 7 \\ 1.5 \\ 9 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 0 \\ 12 \end{pmatrix} = 56 + 0 + 108 = 164\).

The magnitudes are \(|\overrightarrow{OE}| = \sqrt{7^2 + 1.5^2 + 9^2} = \sqrt{132.25} \approx 11.5\) and \(|\overrightarrow{OD}| = \sqrt{8^2 + 0^2 + 12^2} = \sqrt{208}\).

Using the dot product formula, \(164 = 11.5 \times \sqrt{208} \times \cos \theta\).

Solving for \(\theta\), \(\theta = \cos^{-1}\left(\frac{164}{11.5 \times \sqrt{208}}\right) \approx 8.6^\circ\).

(i) To find \(\overrightarrow{AB}\), calculate \(\overrightarrow{OB} - \overrightarrow{OA}\):

\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}.\)

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

The unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{1}{7} \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}.\)

(ii) For \(\angle BOC = 90^\circ\), \(\overrightarrow{OB} \cdot \overrightarrow{OC} = 0.\)

Calculate the dot product:

\(\overrightarrow{OB} \cdot \overrightarrow{OC} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix} = 4 \times 1 + 2 \times 3 + (-2) \times p = 4 + 6 - 2p.\)

Set the dot product to zero:

\(4 + 6 - 2p = 0\)

\(10 = 2p\)

\(p = 5\)

(i) To find the angle between the vectors \(3\mathbf{i} - 4\mathbf{k}\) and \(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\), we use the dot product formula:

\(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\).

The dot product is \((3)(2) + (0)(3) + (-4)(-6) = 6 + 24 = 30\).

The magnitudes are \(|3\mathbf{i} - 4\mathbf{k}| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{25}\) and \(|2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49}\).

Thus, \(30 = \sqrt{25} \times \sqrt{49} \cos \theta\).

Solving for \(\theta\), we get \(\theta = \cos^{-1}\left(\frac{30}{35}\right)\), which is approximately 31° or 0.54 radians.

(ii) \(\overrightarrow{OA}\) is in the same direction as \(3\mathbf{i} - 4\mathbf{k}\) with magnitude 15:

\(\overrightarrow{OA} = (3\mathbf{i} - 4\mathbf{k}) \times \frac{15}{5} = 9\mathbf{i} - 12\mathbf{k}\).

\(\overrightarrow{OB}\) is in the same direction as \(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) with magnitude 14:

\(\overrightarrow{OB} = (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \times \frac{14}{7} = 4\mathbf{i} + 6\mathbf{j} - 12\mathbf{k}\).

(iii) To find the unit vector in the direction of \(\overrightarrow{AB}\), first find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4\mathbf{i} + 6\mathbf{j} - 12\mathbf{k}) - (9\mathbf{i} - 12\mathbf{k}) = -5\mathbf{i} + 6\mathbf{j}\).

The magnitude of \(\overrightarrow{AB}\) is \(\sqrt{(-5)^2 + 6^2} = \sqrt{61}\).

Thus, the unit vector is \(\frac{-5\mathbf{i} + 6\mathbf{j}}{\sqrt{61}}\).

(i) For u to be perpendicular to v, their dot product must be zero:

\(\mathbf{u} \cdot \mathbf{v} = p^2 \cdot 2 + (-2) \cdot (p-1) + 6 \cdot (2p+1) = 0\)

\(2p^2 - 2(p-1) + 6(2p+1) = 0\)

\(2p^2 - 2p + 2 + 12p + 6 = 0\)

\(2p^2 + 10p + 8 = 0\)

Factoring gives:

\((p+1)(p+4) = 0\)

Thus, \(p = -1\) or \(p = -4\).

(ii) For \(p = 1\), calculate the angle between u and v:

\(\mathbf{u} = \begin{pmatrix} 1 \\ -2 \\ 6 \end{pmatrix}, \mathbf{v} = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}\)

\(\mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + (-2) \cdot 0 + 6 \cdot 3 = 20\)

\(|\mathbf{u}| = \sqrt{1^2 + (-2)^2 + 6^2} = \sqrt{41}\)

\(|\mathbf{v}| = \sqrt{2^2 + 0^2 + 3^2} = \sqrt{13}\)

Using the dot product formula:

\(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} = \frac{20}{\sqrt{41} \times \sqrt{13}}\)

\(\theta = \cos^{-1}\left(\frac{20}{\sqrt{41} \times \sqrt{13}}\right) \approx 30.0^\circ\) or \(0.523 \text{ rads}\).

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

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

Calculate the scalar product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (3)(5) + (4)(-2) + (-1)(-3) = 15 - 8 + 3 = 10\).

Calculate the magnitudes:

\(|\overrightarrow{OA}| = \sqrt{3^2 + 4^2 + (-1)^2} = \sqrt{26}\).

\(|\overrightarrow{OB}| = \sqrt{5^2 + (-2)^2 + (-3)^2} = \sqrt{38}\).

Substitute into the scalar product formula:

\(10 = \sqrt{26} \cdot \sqrt{38} \cdot \cos \theta\).

\(\cos \theta = \frac{10}{\sqrt{26} \cdot \sqrt{38}}\).

Calculate \(\theta\):

\(\theta = \cos^{-1}\left(\frac{10}{\sqrt{26} \cdot \sqrt{38}}\right) \approx 71.4\) or \(71.5\) degrees or \(1.25\) radians.

(ii) To find \(\overrightarrow{DC}\), first find \(\overrightarrow{OC}\):

\(\overrightarrow{OC} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \frac{1}{2}((3i + 4j - k) + (5i - 2j - 3k))\).

\(\overrightarrow{OC} = \frac{1}{2}(8i + 2j - 4k) = 4i + j - 2k\).

Given \(\overrightarrow{OD} = 2\overrightarrow{OB} = 2(5i - 2j - 3k) = 10i - 4j - 6k\).

Then \(\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} = (4i + j - 2k) - (10i - 4j - 6k)\).

\(\overrightarrow{DC} = -6i + 5j + 4k\).

(i) To find the value of \(p\) for which angle \(AOB\) is \(90^\circ\), we use the dot product formula. The dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\) when the vectors are perpendicular.

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (5)(2) + (1)(7) + (2)(p) = 10 + 7 + 2p = 0\)

Solving for \(p\):

\(17 + 2p = 0\)

\(2p = -17\)

\(p = -\frac{17}{2}\)

(ii) When \(p = 4\), the vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\mathbf{i} + 7\mathbf{j} + 4\mathbf{k}) - (5\mathbf{i} + \mathbf{j} + 2\mathbf{k})\)

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

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

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

To find the vector with magnitude 28 in the same direction, multiply the unit vector by 28:

\(28 \times \frac{-3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}}{7} = -12\mathbf{i} + 24\mathbf{j} + 8\mathbf{k}\)

(i) The dot product \(\overrightarrow{OA} \cdot \overrightarrow{OB}\) is calculated as:

\((4\mathbf{i} + 7\mathbf{j} - p\mathbf{k}) \cdot (8\mathbf{i} - \mathbf{j} - p\mathbf{k}) = 4 \times 8 + 7 \times (-1) + (-p) \times (-p)\)

\(= 32 - 7 + p^2 = 25 + p^2\)

(ii) For \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) to be perpendicular, their dot product must be zero:

\(25 + p^2 = 0\)

This equation has no real solutions since \(p^2\) cannot be negative.

(iii) For angle AOB to be 60°, we use the cosine formula:

\(\cos 60^{\circ} = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}| |\overrightarrow{OB}|}\)

\(|\overrightarrow{OA}| = \sqrt{4^2 + 7^2 + (-p)^2} = \sqrt{65 + p^2}\)

\(|\overrightarrow{OB}| = \sqrt{8^2 + (-1)^2 + (-p)^2} = \sqrt{65 + p^2}\)

\(\frac{25 + p^2}{65 + p^2} = \frac{1}{2}\)

Solving \(25 + p^2 = \frac{1}{2}(65 + p^2)\) gives:

\(50 + 2p^2 = 65 + p^2\)

\(p^2 = 15\)

\(p = \pm \sqrt{15}\) or \(p = \pm 3.87\)

(a) To find \(\overrightarrow{OC}\), use the property of a parallelogram: \(\overrightarrow{AB} = \overrightarrow{DC}\).

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2\mathbf{i} + 5\mathbf{j} + 3\mathbf{k}) - (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\).

\(\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD}\).

Equating \(\overrightarrow{AB}\) and \(\overrightarrow{DC}\):

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

\(\overrightarrow{OC} = 4\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\).

To verify it's not a rhombus, check if adjacent sides are equal:

\(|\overrightarrow{AB}| = \sqrt{1^2 + 3^2 + 2^2} = \sqrt{14}\).

\(|\overrightarrow{AD}| = \sqrt{(3-1)^2 + (0-2)^2 + (2-1)^2} = \sqrt{6}\).

Since \(\sqrt{14} \neq \sqrt{6}\), it's not a rhombus.

(b) To find \(\angle BAD\), use the dot product:

\(\overrightarrow{BA} = -\overrightarrow{AB} = -\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}\).

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

\(\overrightarrow{BA} \cdot \overrightarrow{AD} = (-1)(2) + (-3)(-2) + (-2)(1) = -2 + 6 - 2 = 2\).

\(|\overrightarrow{BA}| = \sqrt{14}, \quad |\overrightarrow{AD}| = \sqrt{9} = 3\).

\(\cos \angle BAD = \frac{2}{\sqrt{14} \times 3}\).

\(\angle BAD = \cos^{-1}\left(\frac{2}{3\sqrt{14}}\right) \approx 100.3^\circ\).

(c) Area of parallelogram = \(|\overrightarrow{AB} \times \overrightarrow{AD}|\).

\(\overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 2 \\ 2 & -2 & 1 \end{vmatrix}\).

\(= \mathbf{i}(3 \cdot 1 - 2 \cdot (-2)) - \mathbf{j}(1 \cdot 1 - 2 \cdot 2) + \mathbf{k}(1 \cdot (-2) - 3 \cdot 2)\).

\(= \mathbf{i}(3 + 4) - \mathbf{j}(1 - 4) + \mathbf{k}(-2 - 6)\).

\(= 7\mathbf{i} + 3\mathbf{j} - 8\mathbf{k}\).

Magnitude = \(\sqrt{7^2 + 3^2 + (-8)^2} = \sqrt{122} \approx 11.0\).

(i) To find angle \(ABC\), we use the vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\).

\(\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 10 \\ 0 \\ 6 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}\)

The dot product \(\overrightarrow{BA} \cdot \overrightarrow{BC} = (-2)(6) + (1)(-2) + (2)(3) = -8\)

The magnitudes are \(|\overrightarrow{BA}| = \sqrt{(-2)^2 + 1^2 + 2^2} = 3\) and \(|\overrightarrow{BC}| = \sqrt{6^2 + (-2)^2 + 3^2} = 7\).

Using the dot product formula: \(\overrightarrow{BA} \cdot \overrightarrow{BC} = |\overrightarrow{BA}| |\overrightarrow{BC}| \cos \theta\)

\(-8 = 3 \times 7 \times \cos \theta\)

\(\cos \theta = \frac{-8}{21}\)

\(\theta = \cos^{-1}\left(\frac{-8}{21}\right) \approx 112.4^\circ\) or \(1.96\) radians.

(ii) Since ABCD is a parallelogram, \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC}\).

\(\overrightarrow{OD} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \\ 8 \end{pmatrix}\).

(i) First, find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3i + 2j + 8k) - (i - 2j + 4k) = 2i + 4j + 4k\)

\(\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = (3i + 2j + 8k) - (-i - 2j + 10k) = 4i + 4j - 2k\)

Calculate the scalar product \(\overrightarrow{AB} \cdot \overrightarrow{CB}\):

\(\overrightarrow{AB} \cdot \overrightarrow{CB} = (2i + 4j + 4k) \cdot (4i + 4j - 2k) = 2 \times 4 + 4 \times 4 + 4 \times (-2) = 8 + 16 - 8 = 16\)

Find the magnitudes of \(\overrightarrow{AB}\) and \(\overrightarrow{CB}\):

\(|\overrightarrow{AB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\)

\(|\overrightarrow{CB}| = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\)

Use the scalar product to find \(\cos \theta\):

\(\overrightarrow{AB} \cdot \overrightarrow{CB} = |\overrightarrow{AB}| \cdot |\overrightarrow{CB}| \cdot \cos \theta\)

\(16 = 6 \times 6 \times \cos \theta\)

\(\cos \theta = \frac{16}{36} = \frac{4}{9}\)

\(\theta = \cos^{-1}\left(\frac{4}{9}\right) \approx 63.6^\circ\)

(ii) Find the lengths of \(AB\), \(BC\), and \(CA\):

\(AB = |\overrightarrow{AB}| = 6\)

\(BC = |\overrightarrow{CB}| = 6\)

\(CA = |\overrightarrow{CA}| = \sqrt{(-2)^2 + 0^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40}\)

Perimeter of \(\triangle ABC = AB + BC + CA = 6 + 6 + \sqrt{40} \approx 18.32\)

(i) For \(\overrightarrow{OA}\) to be perpendicular to \(\overrightarrow{OB}\), their dot product must be zero:

\((-2)(4) + (3)(1) + (1)(p) = 0\)

\(-8 + 3 + p = 0\)

\(p = 5\)

(ii) The vector \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\) is given by:

\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 1 \\ p \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ p-1 \end{pmatrix}\)

The magnitude of \(\overrightarrow{AB}\) is 7:

\(\sqrt{6^2 + (-2)^2 + (p-1)^2} = 7\)

\(\sqrt{36 + 4 + (p-1)^2} = 7\)

\(36 + 4 + (p-1)^2 = 49\)

\((p-1)^2 = 9\)

\(p-1 = 3\) or \(p-1 = -3\)

\(p = 4\) or \(p = -2\)

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

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

Calculate \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 2 \times 0 + 3 \times (-6) + (-6) \times 8 = -18 - 48 = -66\).

Calculate magnitudes: \(|\overrightarrow{OA}| = \sqrt{2^2 + 3^2 + (-6)^2} = 7\) and \(|\overrightarrow{OB}| = \sqrt{0^2 + (-6)^2 + 8^2} = 10\).

Substitute into the formula: \(-66 = 7 \times 10 \times \cos \theta\).

\(\cos \theta = \frac{-66}{70}\).

\(\theta = \cos^{-1}\left(\frac{-66}{70}\right) = 160.5^\circ\).

(ii) Find \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ 4 \end{pmatrix}\).

Magnitude of \(\overrightarrow{AC} = \sqrt{(-4)^2 + 2^2 + 4^2} = 6\).

To find a vector in the same direction with magnitude 30, scale \(\overrightarrow{AC}\) by \(\frac{30}{6} = 5\).

Vector = \(5 \times \begin{pmatrix} -4 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} -20 \\ 10 \\ 20 \end{pmatrix}\).

(iii) For \(\overrightarrow{OA} + p \overrightarrow{OB}\) to be perpendicular to \(\overrightarrow{OC}\), their dot product must be zero:

\(\left( \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix} + p \begin{pmatrix} 0 \\ -6 \\ 8 \end{pmatrix} \right) \cdot \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} = 0\).

\(\begin{pmatrix} 2 \\ 3 - 6p \\ -6 + 8p \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 5 \\ -2 \end{pmatrix} = 0\).

\(2(-2) + (3 - 6p)5 + (-6 + 8p)(-2) = 0\).

\(-4 + 15 - 30p + 12 - 16p = 0\).

\(23 - 46p = 0\).

\(p = \frac{1}{2}\).

(i) To find \(\overrightarrow{OA} \cdot \overrightarrow{OB}\), calculate the dot product:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (2)(7) + (-8)(2) + (4)(-1) = 14 - 16 - 4 = -6\).

Since the dot product is negative, angle AOB is obtuse.

(ii) First, find \(\overrightarrow{AB}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (7\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (2\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}\).

Then, \(\overrightarrow{AX} = \frac{2}{5} \overrightarrow{AB} = \frac{2}{5} (5\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}) = 2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\).

Now, find \(\overrightarrow{OX}\):

\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = (2\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}) + (2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}) = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\).

Finally, find the unit vector in the direction of \(\overrightarrow{OX}\):

The magnitude of \(\overrightarrow{OX}\) is \(\sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\).

Thus, the unit vector is \(\frac{1}{6} (4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})\).

(i) The vectors \(\mathbf{OA} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{OB} = 3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}\) are perpendicular if their dot product is zero:

\((2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}) = 0\)

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

\(6 - 2 + 2p = 0\)

\(2p = -4\)

\(p = -2\)

(ii) For \(p = 6\), the vectors are \(\mathbf{OA} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{OB} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}\). The dot product is:

\((2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \cdot (3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) = 6 - 2 + 12 = 16\)

The magnitudes are:

\(|\mathbf{OA}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3\)

\(|\mathbf{OB}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{49} = 7\)

Using the cosine formula:

\(16 = 3 \times 7 \times \cos \theta\)

\(\cos \theta = \frac{16}{21}\)

\(\theta \approx 40^\circ\)

(iii) The vector \(\mathbf{AB} = (3\mathbf{i} - 2\mathbf{j} + p\mathbf{k}) - (2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = \mathbf{i} - 3\mathbf{j} + (p - 2)\mathbf{k}\).

The length of \(\mathbf{AB}\) is 3.5:

\(\sqrt{1^2 + (-3)^2 + (p-2)^2} = 3.5\)

\(1 + 9 + (p-2)^2 = 12.25\)

\((p-2)^2 = 2.25\)

\(p-2 = \pm 1.5\)

\(p = 3.5\) or \(p = 0.5\)

(i) First, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}\).

Then, \(\overrightarrow{AC} = 2\overrightarrow{AB} = 2 \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}\).

Now, \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + \begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{7} \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\).

(ii) We have \(\overrightarrow{OD} = m\overrightarrow{OA} + n\overrightarrow{OB}\).

\(m \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + n \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ k \end{pmatrix}\).

This gives the equations:

\(4m + 3n = 1\)

\(m + 2n = 4\)

Solving these simultaneously, we find \(m = -2\) and \(n = 3\).

Substitute \(m\) and \(n\) into the third component equation:

\(-2(-2) + 3(-4) = k\)

\(k = -8\).

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

\(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\)

Given \(\mathbf{a} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}\), the dot product is:

\(\mathbf{a} \cdot \mathbf{b} = (-3)(-1) + (6)(2) + (3)(4) = 3 + 12 + 12 = 27\)

The magnitudes are:

\(|\mathbf{a}| = \sqrt{(-3)^2 + 6^2 + 3^2} = \sqrt{54}\)

\(|\mathbf{b}| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}\)

Thus, \(27 = \sqrt{54} \times \sqrt{21} \times \cos \theta\)

\(\cos \theta = \frac{27}{\sqrt{54} \times \sqrt{21}}\)

\(\theta = \cos^{-1}\left(\frac{27}{\sqrt{54} \times \sqrt{21}}\right) \approx 36.7^\circ\) or \(0.641\) radians

(ii) To find \(\overrightarrow{OC}\), first find \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}\).

Then, \(\overrightarrow{AC} = 3\overrightarrow{AB} = 3 \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix}\).

\(\overrightarrow{OC} = \mathbf{a} + \overrightarrow{AC} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix}\).

The magnitude of \(\overrightarrow{OC}\) is \(\sqrt{3^2 + (-6)^2 + 6^2} = 9\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{9} \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3} \end{pmatrix}\).

Thus, the unit vector is \(\frac{1}{9} \mathbf{i} - \frac{2}{3} \mathbf{j} + \frac{2}{3} \mathbf{k}\).

(i) To find \(\cos POQ\), we use the scalar product formula:

\(\overrightarrow{OP} \cdot \overrightarrow{OQ} = (-2)(2) + (3)(1) + (1)(3) = -4 + 3 + 3 = 2\).

The magnitudes are \(|\overrightarrow{OP}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14}\) and \(|\overrightarrow{OQ}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).

Thus, \(\cos \theta = \frac{\overrightarrow{OP} \cdot \overrightarrow{OQ}}{|\overrightarrow{OP}| |\overrightarrow{OQ}|} = \frac{2}{\sqrt{14} \times \sqrt{14}} = \frac{2}{14} = \frac{1}{7}\).

(ii) The vector \(\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \\ q \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ q-1 \end{pmatrix}\).

The length of \(\overrightarrow{PQ}\) is given by \(\sqrt{4^2 + (-2)^2 + (q-1)^2} = 6\).

Squaring both sides, we have \(16 + 4 + (q-1)^2 = 36\).

Thus, \((q-1)^2 = 16\), giving \(q-1 = 4\) or \(q-1 = -4\).

Therefore, \(q = 5\) or \(q = -3\).

(i) The scalar product \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 8 - 9 - 2 = -3\).

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{14}\) and \(|\overrightarrow{OB}| = \sqrt{29}\).

Using the formula \(\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos \theta\), we have:

\(-3 = \sqrt{14} \times \sqrt{29} \times \cos \theta\)

\(\cos \theta = \frac{-3}{\sqrt{14} \times \sqrt{29}}\)

\(\theta = \cos^{-1}\left(\frac{-3}{\sqrt{14} \times \sqrt{29}}\right) \approx 99^\circ\).

(ii) \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (4 - 2)\mathbf{i} + (-3 - 3)\mathbf{j} + (2 + 1)\mathbf{k} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}\).

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

The unit vector is \(\frac{1}{7}(2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k})\).

(iii) \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = -2\mathbf{i} + 3\mathbf{j} + (p + 1)\mathbf{k}\).

Given \(|\overrightarrow{AB}| = |\overrightarrow{AC}|\), we have:

\(49 = 4 + 9 + (p + 1)^2\)

\(49 = 13 + (p + 1)^2\)

\((p + 1)^2 = 36\)

\(p + 1 = 6\) or \(p + 1 = -6\)

\(p = 5\) or \(p = -7\)

(i) To find \(\overrightarrow{AX}\), calculate \(\overrightarrow{OX} - \overrightarrow{OA} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix}.\)

To show AXB is a straight line, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 10 \\ 2 \\ 11 \end{pmatrix} - \begin{pmatrix} -8 \\ -4 \\ 2 \end{pmatrix} = \begin{pmatrix} 18 \\ 6 \\ 9 \end{pmatrix}.\)

Since \(\overrightarrow{AB} = 3 \overrightarrow{AX}\), AXB is a straight line.

(ii) To show CX is perpendicular to AX, find \(\overrightarrow{CX} = \overrightarrow{OX} - \overrightarrow{OC} = \begin{pmatrix} -2 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -8 \\ 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix}.\)

Calculate the dot product \(\overrightarrow{CX} \cdot \overrightarrow{AX} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix} = -18 + 12 + 6 = 0.\)

Since the dot product is zero, CX is perpendicular to AX.

(iii) To find the area of triangle ABC, calculate the magnitudes \(|\overrightarrow{CX}| = \sqrt{(-3)^2 + 6^2 + 2^2} = 7\) and \(|\overrightarrow{AB}| = \sqrt{18^2 + 6^2 + 9^2} = 21.\)

The area is \(\frac{1}{2} \times |\overrightarrow{CX}| \times |\overrightarrow{AB}| = \frac{1}{2} \times 7 \times 21 = 73.5.\)

(i) The position vectors are \(\mathbf{OA} = \mathbf{i} + 7\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{OB} = -5\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}\).

The dot product \(\mathbf{OA} \cdot \mathbf{OB} = (1)(-5) + (7)(5) + (2)(6) = -5 + 35 + 12 = 42\).

The magnitudes are \(|\mathbf{OA}| = \sqrt{1^2 + 7^2 + 2^2} = \sqrt{54}\) and \(|\mathbf{OB}| = \sqrt{(-5)^2 + 5^2 + 6^2} = \sqrt{86}\).

Using the dot product formula: \(\mathbf{OA} \cdot \mathbf{OB} = |\mathbf{OA}| |\mathbf{OB}| \cos \theta\), we have:

\(42 = \sqrt{54} \sqrt{86} \cos \theta\).

Solving for \(\cos \theta\), we get \(\cos \theta = \frac{42}{\sqrt{54} \sqrt{86}}\).

Thus, \(\theta = \cos^{-1}\left(\frac{42}{\sqrt{54} \sqrt{86}}\right) \approx 0.907\) radians.

(ii) Given \(\overrightarrow{AB} = 2\overrightarrow{BC}\), we find \(\overrightarrow{BC} = \frac{1}{2} \overrightarrow{AB}\).

\(\overrightarrow{AB} = \mathbf{OB} - \mathbf{OA} = (-5\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}) - (\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}) = -6\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\).

Thus, \(\overrightarrow{BC} = \frac{1}{2}(-6\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) = -3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\).

\(\overrightarrow{OC} = \mathbf{OB} + \overrightarrow{BC} = (-5\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}) + (-3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = -8\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}\).

The magnitude of \(\overrightarrow{OC}\) is \(\sqrt{(-8)^2 + 4^2 + 8^2} = \sqrt{144} = 12\).

The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{-8\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}}{12}\).

(i) The vector \(\overrightarrow{AB}\) is given by \(\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\).

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

Thus, the unit vector in the direction of \(\overrightarrow{AB}\) is \(\pm \frac{1}{6} \begin{pmatrix} 2 \\ -4 \\ 4 \end{pmatrix}\).

(ii) For \(\angle AOC = 90^\circ\), the dot product \(\overrightarrow{OA} \cdot \overrightarrow{OC} = 0\).

\(\overrightarrow{OA} \cdot \overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 2 \\ p \end{pmatrix} = 4 + 6 - p = 0\).

Solving for \(p\), we get \(p = 10\).

(iii) The length of \(\overrightarrow{AD}\) is given by \(\sqrt{(-2)^2 + 3^2 + (q+1)^2} = 7\).

\((-2)^2 + 3^2 + (q+1)^2 = 49\).

\(4 + 9 + (q+1)^2 = 49\).

\((q+1)^2 = 36\).

Solving \((q+1)^2 = 36\) gives \(q+1 = 6\) or \(q+1 = -6\).

Thus, \(q = 5\) or \(q = -7\).

(i) Calculate \(\overrightarrow{BA} = \boldsymbol{a} - \boldsymbol{b} = \boldsymbol{i} + 2\boldsymbol{j} - 3\boldsymbol{k}\).

Calculate \(\overrightarrow{BC} = \boldsymbol{c} - \boldsymbol{b} = -2\boldsymbol{i} + 4\boldsymbol{j} + 2\boldsymbol{k}\).

Find the dot product: \((-2) + 8 - 6 = 0\).

Since the dot product is zero, \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are perpendicular.

(ii) Calculate \(\overrightarrow{BC} = \boldsymbol{c} - \boldsymbol{b} = -2\boldsymbol{i} + 4\boldsymbol{j} + 2\boldsymbol{k}\).

Calculate \(\overrightarrow{AD} = \boldsymbol{d} - \boldsymbol{a} = -5\boldsymbol{i} + 10\boldsymbol{j} + 5\boldsymbol{k}\).

These vectors are in the same ratio, hence they are parallel.

Find the ratio of lengths: \(\text{Ratio} = 2:5\) (or \(\sqrt{24} : \sqrt{150}\)).

(i) To find the angle \(\theta\) between \(\mathbf{a}\) and \(\mathbf{b}\), use the dot product formula:

\(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = 2 \times 2 + (-2) \times 6 + 1 \times 3 = 4 - 12 + 3 = -5\).

The magnitudes are \(|\mathbf{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\) and \(|\mathbf{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7\).

Using \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\), we have:

\(-5 = 3 \times 7 \times \cos \theta\).

\(\cos \theta = -\frac{5}{21}\).

\(\theta = \cos^{-1}\left(-\frac{5}{21}\right) \approx 103.8^\circ\) or \(1.81\) radians.

(ii) \(\mathbf{b}\) and \(\mathbf{c}\) are perpendicular if \(\mathbf{b} \cdot \mathbf{c} = 0\).

\(\mathbf{b} \cdot \mathbf{c} = 2p + 6p + 3(p+1) = 11p + 3\).

Set \(11p + 3 = 0\) to find \(p\).

\(11p = -3\).

\(p = -\frac{3}{11}\).

(i) To find the value of \(k\) for which angle \(AOB\) is \(90^\circ\), we use the dot product formula:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = 0\).

\(\begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ k \\ -3 \end{pmatrix} = 6 \times 3 + (-2) \times k + (-6) \times (-3) = 0\).

\(18 - 2k + 18 = 0\).

\(36 - 2k = 0\).

\(k = 18\).

(ii) To find the values of \(k\) for which the lengths of \(OA\) and \(OB\) are equal, we equate their magnitudes:

\(\sqrt{6^2 + (-2)^2 + (-6)^2} = \sqrt{3^2 + k^2 + (-3)^2}\).

\(\sqrt{36 + 4 + 36} = \sqrt{9 + k^2 + 9}\).

\(\sqrt{76} = \sqrt{k^2 + 18}\).

\(76 = k^2 + 18\).

\(k^2 = 58\).

\(k = \pm \sqrt{58}\) or \(k \approx \pm 7.62\).

(iii) For \(k = 4\), find the unit vector in the direction of \(\overrightarrow{OC}\).

\(\overrightarrow{AB} = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} -2 \\ 4 \\ 2 \end{pmatrix}\).

\(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 6 \\ -2 \\ -6 \end{pmatrix} + \begin{pmatrix} -2 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix}\).

The unit vector is \(\frac{1}{6} \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix}\) or \(\frac{1}{6} (4\mathbf{i} + 2\mathbf{j} - 4\mathbf{k})\).

(i) For \(\mathbf{u}\) to be perpendicular to \(\mathbf{v}\), their dot product must be zero:

\(\mathbf{u} \cdot \mathbf{v} = q \cdot 8 + 2 \cdot (q-1) + 6 \cdot (q^2-7) = 0\)

\(8q + 2q - 2 + 6q^2 - 42 = 0\)

\(6q^2 + 10q - 44 = 0\)

Solving this quadratic equation gives \(q = 2\) and \(q = -\frac{11}{3}\).

(ii) When \(q = 0\), \(\mathbf{u} = \begin{pmatrix} 0 \\ 2 \\ 6 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 8 \\ -1 \\ -7 \end{pmatrix}\).

The dot product \(\mathbf{u} \cdot \mathbf{v} = 0 \cdot 8 + 2 \cdot (-1) + 6 \cdot (-7) = -2 - 42 = -44\).

The magnitudes are \(|\mathbf{u}| = \sqrt{0^2 + 2^2 + 6^2} = \sqrt{40}\) and \(|\mathbf{v}| = \sqrt{8^2 + (-1)^2 + (-7)^2} = \sqrt{114}\).

\(\cos \theta = \frac{-44}{\sqrt{40} \times \sqrt{114}} = \frac{-44}{\sqrt{4560}} = \frac{-4}{\sqrt{11}}\).

\(\theta = \cos^{-1}\left(\frac{-4}{\sqrt{11}}\right) \approx 130.7^\circ\) or \(2.28 \text{ radians}\).

(i) To find \(\overrightarrow{AC}\), use the formula \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}\).

\(\overrightarrow{AC} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} - \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix}\).

(ii) The mid-point M of AC is given by \(\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC})\).

\(\overrightarrow{OM} = \frac{1}{2} \left( \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} \right) = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}\).

The unit vector in the direction of \(\overrightarrow{OM}\) is \(\frac{1}{\sqrt{5}} \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}\).

(iii) To find \(\overrightarrow{AB}\), use \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\).

\(\overrightarrow{AB} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}\).

The dot product \(\overrightarrow{AB} \cdot \overrightarrow{AC} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix} = -4 + 24 - 12 = 8\).

The magnitudes are \(|\overrightarrow{AB}| = \sqrt{(-2)^2 + 6^2 + 3^2} = \sqrt{49} = 7\) and \(|\overrightarrow{AC}| = \sqrt{2^2 + 4^2 + (-4)^2} = \sqrt{36} = 6\).

Using the dot product formula, \(\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}| |\overrightarrow{AC}| \cos \theta\), we have \(8 = 7 \times 6 \times \cos \theta\).

Solving for \(\theta\), \(\cos \theta = \frac{8}{42} = \frac{4}{21}\).

\(\theta \approx 79.0^\circ\).

(i) To find \(\overrightarrow{AB}\), calculate \(\overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -10 \\ 3 \\ -13 \end{pmatrix} - \begin{pmatrix} 8 \\ -6 \\ 5 \end{pmatrix} = \begin{pmatrix} -18 \\ 9 \\ -18 \end{pmatrix}\).

The magnitude \(|\overrightarrow{AB}| = \sqrt{(-18)^2 + 9^2 + (-18)^2} = 27\).

To find \(\overrightarrow{BC}\), calculate \(\overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} - \begin{pmatrix} -10 \\ 3 \\ -13 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \\ 12 \end{pmatrix}\).

The magnitude \(|\overrightarrow{BC}| = \sqrt{12^2 + (-6)^2 + 12^2} = 18\).

Since \(|\overrightarrow{AB}|, |\overrightarrow{BC}|, |\overrightarrow{CD}|\) form a geometric progression, \(|\overrightarrow{CD}| = \frac{18}{27} \times 18 = 12\).

(ii) Since \(D\) lies on the line through \(B\) and \(C\), \(\overrightarrow{OD} = \overrightarrow{OB} + t \cdot \overrightarrow{BC}\).

Using the geometric progression, \(\overrightarrow{CD} = \pm \frac{18}{27} \times \overrightarrow{BC}\).

Calculate \(\overrightarrow{OD} = \begin{pmatrix} 2 \\ -3 \\ -1 \end{pmatrix} \pm \frac{18}{27} \times \begin{pmatrix} 12 \\ -6 \\ 12 \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \\ 8 \end{pmatrix}\) or \(\begin{pmatrix} -7 \\ 1 \\ -9 \end{pmatrix}\).