The value of \(\alpha\), so that the volume of the parallelopiped formed by \(\hat{i}+\alpha \hat{j}+\hat{k}, \hat{j}+\alpha \hat{k}\) and \(\alpha \hat{\mathrm{i}}+\hat{\mathrm{k}}\) becomes maximum, is

Volume of parallelopiped is \([\vec{a} \vec{b} \vec{c}]\)
\(\begin{aligned}
\therefore \quad V & =\left|\begin{array}{ccc}
1 & \alpha & 1 \\
0 & 1 & \alpha \\
\alpha & 0 & 1
\end{array}\right| \\
& =1-\alpha\left(-\alpha^2\right)-\alpha \\
& =1+\alpha^3-\alpha
\end{aligned}\)
Differentiating w.r.t. \(\alpha\), we get
\(\begin{aligned}
& \frac{\mathrm{dV}}{\mathrm{d} \alpha}=3 \alpha^2-1 \\
\therefore \quad & \frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \alpha^2}=6 \alpha \\
& \text { Let } \frac{\mathrm{dV}}{\mathrm{d} \alpha}=0 \\
\therefore \quad & 3 \alpha^2-1=0 \\
\therefore \quad & \alpha= \pm \frac{1}{\sqrt{3}} \\
& \text { at } \alpha=\frac{-1}{\sqrt{3}}, \\
& \frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \alpha^2}=\frac{-6}{\sqrt{3}} < 0
\end{aligned}\)
\(\therefore \quad \mathrm{V}\) is maximum at \(\alpha=\frac{-1}{\sqrt{3}}\)