Writing $\vec{c}=\lambda \vec{a}+\mu \vec{b}=(\mu -\lambda )\hat{i}+(\lambda +3\mu )\hat{j}+(3\lambda +\mu )\hat{k}$.

The first dot product gives $3(\mu -\lambda )-6(\lambda +3\mu )+2(3\lambda +\mu )=10$, which simplifies to $-3\lambda -13\mu =10$.

The second dot product gives $(\mu -\lambda )+(\lambda +3\mu )+(3\lambda +\mu )=-2$, which simplifies to $3\lambda +5\mu =-2$.

Adding the two equations yields $-8\mu =8$, so $\mu =-1$, and then $\lambda =1$.

Therefore $\vec{c}=-2\hat{i}-2\hat{j}+2\hat{k}$, and $|\vec{c}{|}^2=4+4+4=12$.