Given that, points $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$
Length $A B=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2+\left(z_A-z_B\right)^2}$
$=\sqrt{(4-2)^2+(7-3)^2+(8-4)^2}$
$=\sqrt{(2)^2+(4)^2+(4)^2}=\sqrt{4+16+16}$
$A B=6$
Length $A C=\sqrt{\left(x_A-x_C\right)^2+\left(y_A-y_C\right)^2+\left(z_A-z_C\right)^2}$
$=\sqrt{(4-2)^2+(7-5)^2+(8-7)^2}$
$=\sqrt{(2)^2+(2)^2+(1)^2}=\sqrt{4+4+1}$
$A C=3 \Rightarrow A B: A C=6: 3=2: 1$
Now, let say the internal bisector of the angle $A$ meets the side $B C$ that point $D\left(x_D, y_D, z_D\right)$
$\therefore x$-coordinate of point $D$,
$x_D=\frac{m x_C+n x_B}{m+n}=\frac{2 \times 2+1 \times 2}{2+1} \Rightarrow x_D=2$
$y$-coordinate of point $D$,
$y_D=\frac{m y_C+n y_B}{m+n}=\frac{2 \times 5+1 \times 3}{2+1}=\frac{13}{3}$
$z$-coordinate of point $D$,
$z_D=\frac{m z_C+n z_B}{m+n}=\frac{2 \times 7+1 \times 4}{2+1}=\frac{18}{3}=6$
$\because$ Point $D\left(2, \frac{13}{3}, 6\right)$
So, length $=\sqrt{(4-2)^2+\left(7-\frac{13}{3}\right)^2+(8-6)^2}$
$=\sqrt{(2)^2+\left(\frac{8}{3}\right)^2+(2)^2}$
$=\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{9}}=\frac{2}{3} \sqrt{34}$