Let a =3 i +2 j + k, b =2 i - j +3 k and c be a vector such that ( a + b ) × c =2( a × b )+24 j -6 k and ( a - b + i ) · c =-3. Then | c |^2 is equal…

Let $\displaystyle \vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\displaystyle \vec{c}$ be a vector such that $\displaystyle (\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $\displaystyle (\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$. Then $\displaystyle |\vec{c}|^2$ is equal to $\displaystyle \_\_\_\_$.