Explanation

(c)
$\overrightarrow{\mathbf{A}}\mathbf{,\ }\overrightarrow{\mathbf{B}}\ and\ \overrightarrow{\mathbf{C}}$ are three vectors, then volume of parallelepiped
$V = \lbrack\ \overrightarrow{\mathbf{A}}\mathbf{\ \ }\overrightarrow{\mathbf{B}}\mathbf{\ \ }\overrightarrow{\mathbf{C}}$ ]
$= \left| \begin{matrix}
1 & a & 1 \\
0 & 1 & a \\
a & 0 & 1
\end{matrix} \right| = 1 + a^{3} - a$
$\Rightarrow V = 1 + a^{3} - a$
On differentiating with respect to $a$, we get
$\frac{dV}{da} = 3a^{2} - 1 = 0$
For maximum or minimum, put $\frac{dV}{a} = 0$
$\Rightarrow a = \pm \frac{1}{\sqrt{3}}$
$\frac{d^{2}V}{da^{2}} = 6a$, positive at $a = \frac{1}{\sqrt{3}}$.
$\therefore V$ is minimum at $a = \frac{1}{\sqrt{3}}$.