If the volume of parallelopiped formed by the vectors $\widehat{\mathrm{i}}+\mathrm{\lambda }\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\widehat{\mathrm{j}}+\mathrm{\lambda }\widehat{\mathrm{k}}\text{ and }\mathrm{\lambda }\widehat{\mathrm{i}}+\widehat{\mathrm{k}}$  is minimum, then 2021$\sqrt{3}$$\mathrm{\lambda }$ is

 

Given vectors are  $\widehat{\mathrm{i}}+\mathrm{\lambda }\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\widehat{\mathrm{j}}+\mathrm{\lambda }\widehat{\mathrm{k}}\text{ and }\mathrm{\lambda }\widehat{\mathrm{i}}+\widehat{\mathrm{k}}$ which forms a parallelopiped.

Volume of the parallelopiped is

$\begin{array}{l}\mathrm{V}=\left|\left|\begin{array}{lll}1& \mathrm{\lambda }& 1\\ 0& 1& \mathrm{\lambda }\\ \mathrm{\lambda }& 0& 1\end{array}\right|\right|=1+{\mathrm{\lambda }}^3-\mathrm{\lambda }\\ \Rightarrow \mathrm{V}={\mathrm{\lambda }}^3-\mathrm{\lambda }+1\end{array}$

On differentiating w.r.t. $\mathrm{\lambda }$, we get

$\frac{\mathrm{dV}}{\mathrm{d\lambda }}=3{\mathrm{\lambda }}^2-1$

For maxima or minima, $\frac{\mathrm{dV}}{\mathrm{d\lambda }}=0\Rightarrow \mathrm{\lambda }=\pm \frac{1}{\sqrt{3}}$

and $\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{d\lambda }}^2}=6\mathrm{\lambda }=\left\{\begin{array}{l}2\sqrt{3}>0,\text{ for }\mathrm{\lambda }=\frac{1}{\sqrt{3}}\\ -2\sqrt{3}<0,\text{ for }\mathrm{\lambda }=-\frac{1}{\sqrt{3}}\end{array}\right.$

$\because \frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{d\lambda }}^2}$ positive for $\mathrm{\lambda }=\frac{1}{\sqrt{3}}$, hence volume V is minimum for $\mathrm{\lambda }=\frac{1}{\sqrt{3}}$