Explanation

(a)
Given that,
$\overrightarrow{\mathbf{a}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{+}3\widehat{j} + 5\widehat{\mathbf{k}}$
and $\overrightarrow{\mathbf{c}}\mathbf{=}7\widehat{\mathbf{i}}\mathbf{+}9\widehat{\mathbf{j}}\mathbf{+}11\widehat{\mathbf{k}}$
Let $\overrightarrow{\mathbf{A}}\mathbf{=}\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{=}\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}} \right) + \left( \widehat{\mathbf{i}} + 3\widehat{\mathbf{j}} + 5\widehat{\mathbf{k}} \right) = 2\widehat{\mathbf{i}} + 4\widehat{\mathbf{j}} + 6\widehat{\mathbf{k}}$
And $\overrightarrow{\mathbf{B}}\mathbf{=}\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}}\mathbf{=}\left( \mathbf{\ }\widehat{\mathbf{i}}\mathbf{+}3\widehat{\mathbf{j}}\mathbf{+}5\widehat{\mathbf{k}} \right) + \left( 7\widehat{\mathbf{i}} + 9\widehat{\mathbf{j}} + 11\widehat{\mathbf{k}} \right) = 8\widehat{\mathbf{i}} + 12\widehat{\mathbf{j}} + 16\widehat{\mathbf{k}}$
If $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ are diagonals, then area of parallelogram
$= \frac{1}{2}\left| \overrightarrow{\mathbf{A}}\mathbf{\times}\overrightarrow{\mathbf{B}}\ \right| = \frac{1}{2}\left\| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
2 & 4 & 6 \\
8 & 12 & 16
\end{matrix} \right\|\$
$= \frac{1}{2}\left| \widehat{\mathbf{i}}(64 - 72) - \widehat{\mathbf{j}}(32 - 48) + \widehat{\mathbf{k}}(24 - 32) \right|\$
$= \frac{1}{2}\left| - 8\widehat{\mathbf{i}} + 16\widehat{\mathbf{j}} - 8\widehat{\mathbf{k}}\ \right|$
$= \left| - 4\widehat{\mathbf{i}} + 8\widehat{\mathbf{j}} - 4\widehat{\mathbf{k}} \right|$
$= \sqrt{( - 4)^{2} + (8)^{2} + ( - 4)^{2}}$
$= \sqrt{16 + 64 + 16} = \sqrt{96} = 4\sqrt{6}$