Explanation

(a)
Let $\overrightarrow{\mathbf{A}} = \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} = \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) + \left( \widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 5\widehat{\mathbf{k}} \right)$
$= 2\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$
$and\ \overrightarrow{\mathbf{B}} = \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = \left( \widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 5\widehat{\mathbf{k}} \right) + \left( 7\widehat{\dot{\mathbf{i}}} + 9\widehat{\dot{\mathbf{j}}} + 11\widehat{\mathbf{k}} \right)$
$= 8\widehat{\dot{\mathbf{i}}} + 12\widehat{\dot{\mathbf{j}}} + 16\widehat{\mathbf{k}}$
$\therefore Area\ of\ parallelogram = \frac{1}{2}||\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}||$
$= \frac{1}{2}\left\| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
2 & 4 & 6 \\
8 & 12 & 16
\end{matrix} \right\|$
$= \frac{1}{2}| - 8\widehat{\dot{\mathbf{i}}} + 16\widehat{\dot{\mathbf{j}}} - 8\widehat{\mathbf{k}}|$
$= \sqrt{{( - 4)}^{2} + {(8)}^{2} + {( - 4)}^{2}}$
$= 4\sqrt{6}$ sq units