Let $\alpha \beta \gamma = 45$; $\alpha, \beta, \gamma \in R$. If $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in R$ such that $xyz \neq 0$,then $6\alpha + 4\beta + \gamma$ is equal to:

  • A
    $55$
  • B
    $56$
  • C
    $54$
  • D
    $31$

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