If $\bar{a}=\frac{1}{\sqrt{10}}(3 \hat{\imath}+\hat{k})$ and $\bar{b}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a}+2 \bar{b})]$ is

  • A
    $7$
  • B
    $-5$
  • C
    $5$
  • D
    $-7$

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