In $∆\mathrm{ABC}$, A(4, 7, 8), B(2, 3, 4), C(2, 5, 7) then the length of the internal bisector of A is

The internal angular bisector from A divides BC in the ratio AB:AC

Hence 

$$\begin{gathered} AB=\sqrt{4+16+16}=6 \\ AC=\sqrt{4+4+1}=3 \end{gathered}$$

Hence, the internal  angular bisector of Angle $A$ divides $BC$ in the ratio $2:1$ internally

Hence, $D=\left(\frac{4+2}{3},\frac{10+3}{3},\frac{14+4}{3}\right)=\left(2,\frac{13}{3},6\right)$

The required distance is AD

$\begin{array}{rcl}AD& =& \sqrt{4+{\left(7-\frac{13}{3}\right)}^2+4}\\ & =& \sqrt{8+\frac{64}{9}}\\ & =& \frac{\sqrt{72+64}}{3}\\ & =& \frac{2\sqrt{34}}{3}\end{array}$