Let a → = 3 i ^ + 2 j ^ + k ^ , b → = 2 i ^ − j ^ + 3 k ^ and c → be a vector such that a → + b → × c → = 2…

Given,

$\overrightarrow{a}=3\hat{i}+2\hat{j}+\hat{k}, \overrightarrow{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\overrightarrow{c}$ be a vector such that $\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \overrightarrow{c}=2\left(\overrightarrow{a}\times \overrightarrow{b}\right)+24\hat{j}-6\hat{k}$ 

Now, finding $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 1\\ 2& -1& 3\end{array}\right|=7\hat{i}-7\hat{j}-7\hat{k}$

Now, solving

$\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \overrightarrow{c}=2\left(\overrightarrow{a}\times \overrightarrow{b}\right)+24\hat{j}-6\hat{k}$

$\Rightarrow \left(5\hat{i}+\hat{j}+4\hat{k}\right)\times \overrightarrow{c}=2\left(7\hat{i}-7\hat{j}-7\hat{k}\right)+24\hat{j}-6\hat{k}$

$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 5& 1& 4\\ x& y& z\end{array}\right|=14\hat{i}+10\hat{j}-20\hat{k}$

$\Rightarrow \hat{i}\left(z-4y\right)-\hat{j}\left(5z-4x\right)+\hat{k}\left(5y-x\right)=14\hat{i}+10\hat{j}-20\hat{k}$

Now, on comparing both side, 

$z-4y=14, 4x-5z=10, 5y-x=-20 ....\left(i\right)$

Now, solving 

$\left(\overrightarrow{a}-\overrightarrow{b}+\hat{i}\right)·\overrightarrow{c}=-3$

$\Rightarrow \left(2\hat{i}+3\hat{j}-2\hat{k}\right)·\overrightarrow{c}=-3$

$\Rightarrow 2x+3y-2z=-3 .......\left(ii\right)$

Now, solving the equation $\left(i\right) \& \left(ii\right)$ we get,

$\therefore x=5, y=-3, z=2$

$\Rightarrow {\left|\overrightarrow{c}\right|}^2=25+9+4=38$