If $\bar{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k})$,$\bar{b}=(-\hat{i}+2 \hat{j}+\hat{k})$ and $\bar{c}=(3 \hat{i}+\hat{j})$ such that $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

  • A
    -$8$
  • B
    $8$
  • C
    $10$
  • D
    $\frac{8}{3}$

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