Explanation

(d)
We have, $\overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{c}}$
$\Rightarrow \ \overrightarrow{\mathbf{c}}$ is perpendicular to $\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{b}}\mathbf{\times}\overrightarrow{\mathbf{c}}\mathbf{=}\overrightarrow{\mathbf{a}}$.
$\Rightarrow \ \overrightarrow{\mathbf{a}}$ is perpendicular to $\overrightarrow{\mathbf{b}}\ and\ \overrightarrow{\mathbf{c}}$.
$\Rightarrow \ \overrightarrow{\mathbf{a}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{,}\overrightarrow{\mathbf{c}}$ are mutually perpendicular.
Again $\overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{c}}$
$\Rightarrow \left| \overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}} \right|\mathbf{= |}\overrightarrow{\mathbf{c}}|$
$= |\overrightarrow{\mathbf{a}}\mathbf{||}\overrightarrow{\mathbf{b}}| \bullet \sin{90{^\circ} =}|\overrightarrow{\mathbf{c}}|$
$\Rightarrow \left| \overrightarrow{\mathbf{a}} \right|\left| \overrightarrow{\mathbf{b}} \right|\mathbf{= |}\overrightarrow{\mathbf{c}}|\$ ...(i)
Also, $\overrightarrow{\mathbf{b}}\mathbf{\times}\overrightarrow{\mathbf{c}}\mathbf{= |}\overrightarrow{\mathbf{a}}|$
$\left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right| \bullet \sin{90{^\circ} =}|\overrightarrow{\mathbf{a}}|$
$\left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right|\mathbf{=}\mathbf{|}\overrightarrow{\mathbf{a}}|$ ….(ii)
From Eqs. (i) and (ii), we get
$\left| \overrightarrow{\mathbf{b}} \right|^{2}\left| \overrightarrow{\mathbf{c}} \right| = |\overrightarrow{\mathbf{c}}|$
$\therefore\ \left| \overrightarrow{\mathbf{b}} \right|^{2} = 1\ \ (\because\ |\overrightarrow{\mathbf{c}}| \neq 0)$
$\Rightarrow \ \left| \overrightarrow{\mathbf{b}} \right| = 1$
$\Rightarrow \ \left| \overrightarrow{\mathbf{a}} \right|\mathbf{= |}\overrightarrow{\mathbf{c}}|$