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