If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$,$\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ and $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$,then the value of $x$ is:

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

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