Explanation

(b)
Let , $\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}},\ \ \overrightarrow{\mathbf{b}} = - \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}}$
$Now,\ \ \ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
1 & - 3 & 2 \\
- 1 & 2 & 0
\end{matrix} \right| = - 4\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
$\therefore Area\ of\ parallelogram = \frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
$= \frac{1}{2}\sqrt{16 + 4 + 1} = \frac{\sqrt{21}}{2}$