Explanation

(d)
If $\overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}}$ is collinear with $\overrightarrow{\mathbf{c}}$, then
$\overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}} = t\overrightarrow{\mathbf{c}}$ ….(i)
Also, if $\overrightarrow{\mathbf{b}} + 3\overrightarrow{\mathbf{c}}$ is collinear with $\overrightarrow{\mathbf{a}}$ then
$\overrightarrow{\mathbf{b}} + 3\overrightarrow{\mathbf{c}} = \lambda\overrightarrow{\mathbf{a}}$
$\Longrightarrow \ \overrightarrow{\mathbf{b}} = \lambda\overrightarrow{\mathbf{a}} - 3\overrightarrow{\mathbf{c}}$ …(ii)
On putting the value of $\overrightarrow{\mathbf{b}}$ in Eq. (i), we get
$\overrightarrow{\mathbf{a}} + 2\left( \lambda\overrightarrow{\mathbf{a}} - 3\overrightarrow{\mathbf{c}} \right) = t\overrightarrow{\mathbf{c}}$
$\Longrightarrow \left( \overrightarrow{\mathbf{a}} - 6\overrightarrow{\mathbf{c}} \right) = t\overrightarrow{\mathbf{c}} - 2\lambda\ \overrightarrow{\mathbf{a}}$
On comparing, we get $1 = - 2\lambda\ and - 6 = t$
$\Longrightarrow \ \lambda = - \frac{1}{2}\ and\ t = - 6$
From Eq. (i)
$\overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}} = - 6\overrightarrow{\mathbf{c}}$
$\mathbf{\Longrightarrow \ }\overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}} + 6\overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}}$