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If $\mathrm{A}(4,7,8), \mathrm{B}(2,3,4)$ and $\mathrm{C}(2,5,7)$ are the position vectors of the vertices of a triangle $\mathrm{ABC}$ and if the internal bisector of $\angle A$ meets $\mathrm{BC}$ at $\mathrm{D}$ then $\mathrm{AD}=$
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The correct answer is:
$\frac{2}{3} \sqrt{34}$
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