Given: $A(4,7,8),\:B(2,3,4)\:and\:C(2,5,7)$
$\Rightarrow\:AB=\sqrt {4+16+16}=6$
$\Rightarrow\:AC=\sqrt {4+4+1}=3$
Let the angular bisector of $A$ be $AD$, where $D$ is on $BC$.
$\Rightarrow\:D$ divides $BC$ in the ratio $6:3=2:1$
$\therefore $ coordinates of $D$ using section formula is given by
$D\big(\large\frac{4+2}{2+1},\frac{10+3}{2+1},\frac{14+4}{2+1}\big)$$=(2,\large\frac{13}{3}$$,6)$
Now
Th length of the internal angular bisector of angle $A$ is $AD$
$AD=\sqrt {4+64/9+4}=\large\frac{\sqrt {136}}{3}=\frac{2\sqrt {34}}{3}$