Explanation

(b)
Given, $2\overrightarrow{\mathbf{a}} + 3\overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}}$
$\Longrightarrow 2\overrightarrow{\mathbf{a}} + 3\overrightarrow{\mathbf{b}} = - \overrightarrow{\mathbf{c}}$
Taking cross product with $\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}$ respectively, we get
$2\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}} \right) + 3\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) = - \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}$
$\Longrightarrow 3\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) = - \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$ …(i)
$and\ 2\left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}} \right) + 3\left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}} \right) = - \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$
$\Longrightarrow 2\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$ ….(ii)
Now, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$
$= \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$+3($\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$ [using Eq. (i)]
$= 4\left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) + \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$
$= 2\left( \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} \right) + \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$ [using Eq. (ii)]
$= 3(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$