Explanation

(d)
Since,
$\left\lbrack 3\overrightarrow{\mathbf{v}}\ p\ \overrightarrow{\mathbf{v}}\ p\ \overrightarrow{\mathbf{w}} \right\rbrack - \left\lbrack p\ \overrightarrow{\mathbf{v}}\ \overrightarrow{\mathbf{w}}\ q\ \overrightarrow{\mathbf{u}} \right\rbrack - \left\lbrack 2\ \overrightarrow{\mathbf{w}}\ q\ \overrightarrow{\mathbf{v}}\ q\ \overrightarrow{\mathbf{u}} \right\rbrack = 0$
$\therefore 3p^{2}\left\lbrack \overrightarrow{\mathbf{u}} \bullet \left( \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{w}} \right) \right\rbrack - pq\left\lbrack \overrightarrow{\mathbf{v}}\ \bullet \left( \overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}} \right) \right\rbrack$
$- 2q^{2}\left\lbrack \overrightarrow{\mathbf{w}} \bullet \left( \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{u}} \right) \right\rbrack = 0$
$\Longrightarrow \left( 3p^{2} - pq + 2q^{2} \right)\left\lbrack \overrightarrow{\mathbf{u}} \bullet \left( \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{w}} \right) \right\rbrack = 0$
But $\left\lbrack \overrightarrow{\mathbf{u}}\ \overrightarrow{\mathbf{v}}\ \overrightarrow{\mathbf{w}} \right\rbrack \neq 0$
$\Longrightarrow 3p^{2} - pq + {2q}^{2} = 0$
$\Longrightarrow p = q = 0$