If the area of parallelogram, whose adjacent sides are $3\widehat{i}+4\widehat{j}-\lambda \widehat{k}$ and $2\widehat{j}-4\widehat{k}$ is $\sqrt{436}$ sq unit, then $\lambda$ is equal to

Area of parallelogram $=\left|\overrightarrow{a}\times \overrightarrow{b}\right|$
Given, $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{436}$
Now, $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 3& 4& -\lambda \\ 0& 2& -4\end{array}\right|$

$\begin{array}{l}=\widehat{i}\left(-16+2\lambda \right)-\widehat{j}\left(-12-0\right)+\widehat{k}\left(6\right)\\ =2\left(\lambda -8\right)\widehat{i}+12\widehat{j}+6\widehat{k}\end{array}$

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{4{\left(\lambda -8\right)}^2+144+36}$

$=2\sqrt{{\left(\lambda -8\right)}^2+36+9}$

$\Rightarrow \sqrt{436}=2\sqrt{{\left(\lambda -8\right)}^2+45}$
$\Rightarrow 436=4\left[{\lambda }^2-16\lambda +64+45\right]$

$\Rightarrow 109={\lambda }^2-16\lambda +109$

$\Rightarrow {\lambda }^2-16\lambda =0$

$\Rightarrow \lambda =16,0$