Explanation

(d)
Given, $\left| \overrightarrow{\mathbf{a}} \right| = 1,\ |\overrightarrow{\mathbf{b}}| = 2$
$\therefore\lbrack(\overrightarrow{\mathbf{a}} + 3\overrightarrow{\mathbf{b}}) \times (3\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}})\rbrack^{2}$
$= \lbrack 0 + \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} + 9\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}} + 0\rbrack^{2}$
$= \lbrack - 8\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}\rbrack^{2}$
$= 64\left\lbrack \left| \overrightarrow{\mathbf{a}} \right|^{2}\ \middle| \overrightarrow{\mathbf{b}}|^{2}\sin^{2}\theta \right\rbrack$
$= 64\left\lbrack 1 \times 4 \times \sin^{2}{120{^\circ}} \right\rbrack$
$= 64 \times 4 \times \frac{3}{4} = 192$