If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$. Then the triplet $(\alpha, \beta, \gamma)$ is

  • A
    $(2, -1, -1)$
  • B
    $(-2, 1, 1)$
  • C
    $(-2, -1, 1)$
  • D
    $(2, 1, -1)$

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