If the position vectors of the vertices $A, B$ and $C$ of $\triangle ABC$ are $\hat{i}+2\hat{j}-5\hat{k}$,$-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively,then $\angle B=$

  • A
    $\cos^{-1}\left(\frac{7}{3\sqrt{10}}\right)$
  • B
    $\cos^{-1}\left(\frac{8}{\sqrt{105}}\right)$
  • C
    $\cos^{-1}\left(\frac{1}{\sqrt{42}}\right)$
  • D
    $\cos^{-1}\left(-\frac{7}{3\sqrt{10}}\right)$

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