Explanation

(d)
Given,
$\frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}||\overrightarrow{\mathbf{a}} = (\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}$
$\therefore\ \frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}||\overrightarrow{\mathbf{a}} = (\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{c}})\overrightarrow{\mathbf{b}} - (\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}})\overrightarrow{\mathbf{a}}$
On comparing the coefficient of $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, we get
$\frac{1}{2}\left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right| = - \overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}}\ and\ \overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{c}} = 0$
$\Longrightarrow \frac{1}{3}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}| = - |\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{c}}|cos\ \theta \Longrightarrow \cos{\theta = - \frac{1}{3}}$
$\Longrightarrow 1 - \sin^{2}\theta = \frac{1}{9} \Longrightarrow \sin\theta = \frac{2\sqrt{2}}{3}$