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 respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is

  • A
    $2 \hat{i}+\frac{13}{3} \hat{j}+2 \hat{k}$
  • B
    $2 \hat{i}-\frac{13}{3} \hat{j}+6 \hat{k}$
  • C
    $2 \hat{i}+13 \hat{j}+6 \hat{k}$
  • D
    $2 \hat{i}+\frac{13}{3} \hat{j}+6 \hat{k}$

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