Let $\overline{a} = \hat{i} + \hat{j} + \hat{k}$,$\overline{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overline{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$. If the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,then $x = \dots$

  • A
    $0$
  • B
    $1$
  • C
    $-4$
  • D
    $-2$

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