If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = 0$,then $\lambda + \mu =$ . . . . . . .

  • A
    $-\frac{21}{2}$
  • B
    $\frac{23}{2}$
  • C
    $\frac{33}{2}$
  • D
    $33$

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