Explanation

(d)
Since, $\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}\$is collinear to $\overrightarrow{\mathbf{C}}$ and $\overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}}$ is collinear to $\overrightarrow{\mathbf{A}}$
$\therefore\ \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} = \lambda\ \overrightarrow{\mathbf{C}}\ and\ \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} = \mu\ \overrightarrow{\mathbf{A}}$
Where λ and $\mu$ are scalars.
$\Longrightarrow \ \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} = (\lambda + 1)\overrightarrow{\mathbf{C}}$
and $\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} = (\mu + 1)\overrightarrow{\mathbf{A}}$
$\Longrightarrow (\lambda + 1)\overrightarrow{\mathbf{C}} = (\mu + 1)\overrightarrow{\mathbf{A}}$
If λ$\neq - 1$, then
$\overrightarrow{\mathbf{C}} = \frac{\mu + 1}{\lambda + 1}\overrightarrow{\mathbf{A}}$
⟹ $\overrightarrow{\mathbf{C}}\ and\ \overrightarrow{\mathbf{A}}\$are collinear.
This is a contradiction to the given condition.
$\therefore\ \ \lambda = - 1$
$\therefore\ \ \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{0}}$