The Volume of parallelopiped is given by $\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]$, where $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ represents the vectors that form the parallelopiped.

Putting the given vectors, we get

$\Rightarrow V=\left|\begin{array}{ccc}1& a& 1\\ 0& 1& a\\ a& 0& 1\end{array}\right|$

$=1-a\left(-a^2\right)-a$

$=1+a^3-a$

For minimum value $\frac{dV}{da}=0$

$\Rightarrow \frac{dV}{da}=3a^2-1=0$

$\Rightarrow a=\pm \frac{1}{\sqrt{3}}$

Now, at $a=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{d^{2}V}{d{a}^2}=6a$

$\Rightarrow \frac{d^{2}V}{d{a}^2}=\frac{6}{\sqrt{3}}>0$

Therefore, Volume $V$ is minimum at $a=\frac{1}{\sqrt{3}}$.