(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, \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC}\).

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