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

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

And  $\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{0}$,

Let $\overrightarrow{c}=x\hat{i}+y\hat{j}+z\hat{k}$

So,  $\overrightarrow{a}\times \left(\overrightarrow{b}+\overrightarrow{c}\right)=\overrightarrow{0}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{0}$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{a}\times \overrightarrow{c}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}$

$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 1& 1& 1\end{array}\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ 1& -2& 3\end{array}\right|$

$\Rightarrow -5\hat{i}+2\hat{j}+3\hat{k}=\hat{i}\left(3y+2z\right)+\hat{j}\left(z-3x\right)+\hat{k}\left(-2x-y\right)$

On comparing we get,

$3y+2z=-5$ .....(i)

$z-3x=2$ ............(ii)

$-2x-y=3$ ........(iii)

On solving equation (i), (ii) and (iii) we get,

$x=\frac{-1}{6}, y=\frac{-8}{3}, z=\frac{3}{2}$

So, $\overrightarrow{c}=\frac{-1}{6}\hat{i}+\frac{-8}{3}\hat{j}+\frac{3}{2}\hat{k}$

So, $\overrightarrow{c}.\overrightarrow{a}=\left(\frac{-1}{6}\hat{i}+\frac{-8}{3}\hat{j}+\frac{3}{2}\hat{k}\right).\left(\hat{i}-2\hat{j}+3\hat{k}\right)=\frac{29}{3}$

So, $3\overrightarrow{c}.\overrightarrow{a}=29$

Note: This question appeared in JEE Main 2022, 29th june shift 2. The question was incorrect, so it is modified to make it correct.