Let a = i - j, b = i + j + k and c be a vector such that a × c + b = 0 and a · c =4, then | c |^2 is equal to:

Let $\displaystyle \vec{a}=\hat{i}-\hat{j}, \vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\displaystyle \vec{c}$ be a vector such that $\displaystyle \vec{a} \times \vec{c}+\vec{b}=\overrightarrow{0}$ and $\displaystyle \vec{a} \cdot \vec{c}=4$, then $\displaystyle |\vec{c}|^2$ is equal to: