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