Solve this following

Question:

If the position vectors of the vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a $\triangle \mathrm{ABC}$ are respectively

$4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}, 2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$, then the position vector of the point, where the bisector of $\angle \mathrm{A}$ meets $\mathrm{BC}$ is :-

 

  1. $\frac{1}{2}(4 \hat{i}+8 \hat{j}+11 \hat{k})$

  2. $\frac{1}{3}(6 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}+18 \hat{\mathrm{k}})$

  3. $\frac{1}{4}(8 \hat{\mathrm{i}}+14 \hat{\mathrm{j}}+19 \hat{\mathrm{k}})$

  4. $\frac{1}{3}(6 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+15 \hat{\mathrm{k}})$


Correct Option: , 2

Solution:

Solution not required 

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