If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

  • A
    $\frac{1}{\sqrt{13}+6} (4\sqrt{13}+12)\hat{i} + (7\sqrt{13}+30)\hat{j} + (8\sqrt{13}+42)\hat{k}$
  • B
    $\frac{1}{\sqrt{13}-6} (4\sqrt{13}+12)\hat{i} + (7\sqrt{13}+30)\hat{j} + (8\sqrt{13}+42)\hat{k}$
  • C
    $\frac{1}{\sqrt{13}+6} (4\sqrt{13}+12)\hat{i} + (7\sqrt{13}+30)\hat{j} + (8\sqrt{13}+42)\hat{k}$
  • D
    $\frac{1}{6-\sqrt{13}} (4\sqrt{13}+12)\hat{i} + (7\sqrt{13}+30)\hat{j} - (8\sqrt{13}+42)\hat{k}$

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