Explanation

(a)
Vector perpendicular to face $OAB$ is ${\overrightarrow{\mathbf{n}}}_{1}$
$= \overrightarrow{\mathbf{OA}} \times \overrightarrow{\mathbf{OB}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
1 & 2 & 1 \\
2 & 1 & 3
\end{matrix} \right|$
$= 5\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}}$
Vector perpendicular to face $ABC$ is${\overrightarrow{\mathbf{n}}}_{2}$
$= \overrightarrow{\mathbf{AB}} \times \overrightarrow{\mathbf{AC}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
1 & - 1 & 2 \\
- 2 & - 1 & 1
\end{matrix} \right|$
$= \widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}}$
$\therefore\cos{\theta = \frac{{\overrightarrow{\mathbf{n}}}_{1} \bullet {\overrightarrow{\mathbf{n}}}_{2}}{\left| {\overrightarrow{\mathbf{n}}}_{1} \right|\left| {\overrightarrow{\mathbf{n}}}_{2} \right|}}$
$= \frac{5 \times 1 + ( - 1) \times ( - 5) + ( - 3) \times ( - 3)}{\sqrt{5^{2} + ( - 1)^{2} + ( - 3)^{2}}\sqrt{1^{2} + ( - 5)^{2} + ( - 3)^{2}}}$
$= \frac{5 + 5 + 9}{\sqrt{35}\sqrt{35}} = \frac{19}{35}$
$\Longrightarrow \ \theta = \cos^{- 1}\left( \frac{19}{35} \right)$