Let λ ∈ ℝ , a → = λ i ^ + 2 j ^ - 3 k ^ , b → = i ^ - λ j ^ + 2 k ^ ,…

Given,

$\overrightarrow{a}=\lambda \hat{\mathit{i}}+2\hat{j}-3\hat{\mathit{k}}$

$\overrightarrow{b}=\hat{\mathit{i}}-\lambda \hat{\mathrm{j}}+2\hat{\mathit{k}}$

Also given,

$\left(\right(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}\times \overrightarrow{b}\left)\right)\times (\overrightarrow{a}-\overrightarrow{b})=8\hat{i}-40\hat{j}-24\hat{k}$

$\Rightarrow (\overrightarrow{b}-\overrightarrow{a})\times \left(\right(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}\times \overrightarrow{b}\left)\right)=8\hat{i}-40\hat{j}-24\hat{k}$

Now using vector triple product,

$\overrightarrow{A}\times (\overrightarrow{B}\times \overrightarrow{C})=(\overrightarrow{A}.\overrightarrow{C})\overrightarrow{B}-(\overrightarrow{A}.\overrightarrow{B})\overrightarrow{C}$ we get,

$\Rightarrow \left(\right(\overrightarrow{b}-\overrightarrow{a})·(\overrightarrow{a}\times \overrightarrow{b}\left)\right)(\overrightarrow{a}+\overrightarrow{b})-\left(\right(\overrightarrow{b}-\overrightarrow{a})·(\overrightarrow{a}+\overrightarrow{b}\left)\right)(\overrightarrow{a}\times \overrightarrow{b})=8\hat{i}-40\hat{j}-24\hat{k}$

$\Rightarrow 0+\left(\right(\overrightarrow{a}-\overrightarrow{b})·(\overrightarrow{a}+\overrightarrow{b}\left)\right)(\overrightarrow{a}\times \overrightarrow{b})=8\hat{i}-40\hat{j}-24\hat{k}.....\left(1\right)$

$\left(\right(\overrightarrow{a}-\overrightarrow{b})·(\overrightarrow{a}+\overrightarrow{b}\left)\right)=\left[\left(\lambda -1\right)\hat{i}+\left(2+\lambda \right)\hat{j}-5\hat{k}\right].\left[\left(\lambda +1\right)\hat{i}+\left(2-\lambda \right)\hat{j}-\hat{k}\right]$

$\Rightarrow \left(\right(\overrightarrow{a}-\overrightarrow{b})·(\overrightarrow{a}+\overrightarrow{b}\left)\right)=8$

From equation (1) we get,

$\Rightarrow 8(\overrightarrow{a}\times \overrightarrow{\mathit{b}})=8\hat{i}-40\hat{\mathit{j}}-24\hat{\mathit{k}}$

Now, $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \lambda & 2& -3\\ 1& -\lambda & 2\end{array}\right|$

$=(4-3\lambda )\hat{i}-(2\lambda +3)\hat{j}+\left(-{\lambda }^2-2\right)\hat{k}$

$\Rightarrow \lambda =1$

$\therefore \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}$

$\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}$

$\Rightarrow \overrightarrow{a}+\overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k},\overrightarrow{a}-\overrightarrow{b}=3\hat{j}-5\hat{k}$

$\Rightarrow (\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 0& 3& -5\end{array}\right|=2\hat{i}+10\hat{j}+6\hat{k}$

$\Rightarrow \left|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})\right|=\sqrt{4+100+36}$

Now,

${\left|\lambda (\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})\right|}^2={\lambda }^2{\left|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})\right|}^2$

$\therefore$ Required answer $=4+100+36=140$