$D(1,5,-1), E(0,4,-2), F(2,3,4)$ are mid-points
Then,
$$
\begin{aligned}
& \frac{x_1+x_2}{2}=1 \frac{y_1+y_2}{2}=5 \frac{z_1+z_2}{2}=-1 \\
& x_1+x_2=2, \quad y_1+y_2=10, \quad z_1+z_2=-2
\end{aligned}
$$

Similarly,
$$
\begin{array}{lll}
x_2+x_3=0, & y_2+y_3=8, & z_2+z_3=-4 \\
x_1+x_3=4, & y_1+y_3=6, & z_1+z_3=8
\end{array}
$$

On adding, $2\left(x_1+x_2+x_3\right)=6$,
$$
\begin{aligned}
& 2\left(y_1+y_2+y_3\right)=24, \quad 2\left(z_1+z_2+z_3\right)=2 \\
& x_1+x_2+x_3=3, \quad y_1+y_2+y_3=12 \\
& z_1+z_2+z_3=1 \\
& \Rightarrow \quad x_3=1, y_3=2, z_3=3 \\
&
\end{aligned}
$$
length of median drawn from $C$ to $A B$ is
$$
\begin{aligned}
C D & =\sqrt{(1-1)^2+(2-5)^2+(-1-3)^2} \\
& =\sqrt{0+9+16}=\sqrt{25}=5
\end{aligned}
$$