If the volume of parallelopiped formed by the vectors $\overline{i}+\lambda \overline{j}+\overline{k},\overline{j}+\lambda \overline{k}$   and  $\lambda \overline{i}+\overline{k}$  is minimum then  $\left(2025\right){\lambda }^2=\_\_\_\_$

Volume V $=\left|\begin{array}{ccc}1& \lambda & 1\\ 0& 1& \lambda \\ \lambda & 0& 1\end{array}\right|=1+{\lambda }^3-\lambda$
So  $f\left(\lambda \right)={\lambda }^3-\lambda +1 f\text{'}\left(\lambda \right)=3{\lambda }^2-1,f^{\text{'}\text{'}}\left(\lambda \right)=6\lambda$
      
For maxima (or) minima   $f\text{'}\left(\lambda \right)=0\Rightarrow 3{\lambda }^2-1=0\Rightarrow \lambda =\pm \frac{1}{\sqrt{3}}$
at $\lambda =\frac{-1}{\sqrt{3}}\Rightarrow f\text{'}\text{'}\left(\lambda \right)=\frac{-6}{\sqrt{3}}<0\Rightarrow$  maximum 
at $\lambda =\frac{1}{\sqrt{3}}\Rightarrow f\text{'}\text{'}\left(\lambda \right)=\frac{6}{\sqrt{3}}>0\Rightarrow$  minimum 
$\therefore \left(2025\right){\lambda }^2=675$