If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$ are three vectors such that $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then $x = ...$
- A$0$
- B$1$
- C$-4$
- D$-2$
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