Explanation

(b)
Given $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}}$
$\Longrightarrow \ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}} = 0$
$\Longrightarrow \ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$
Similarly, $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}$
$\therefore\ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} \neq \overrightarrow{\mathbf{0}}$
Alternate: Since, $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ are unit vectors and $\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{0}},$
so $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}},\overrightarrow{\mathbf{c}}$ represent an equilateral triangle.
$\mathbf{\therefore\ \ }\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}} \neq \overrightarrow{\mathbf{0}}$