$\therefore(l, 0,0)$ is mid point of AB.
$\Rightarrow x_1+x_2=2 l ; y_1+y_2=0 ; z_1+z_2=0$ ...(i)
$(0, m, 0)$ is mid point of $\mathrm{BC}$.
$\therefore\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}\right)=(0, \mathrm{~m}, 0)$
$\Rightarrow x_2+x_3=0, y_2+y_3=2 m, z_2+z_3=0$ ...(ii)
Since $(0,0, n)$ is mid point of AC.
$\therefore\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}, \frac{z_1+z_3}{2}\right)=(0,0, n)$
$\Rightarrow x_1+x_3=0, y_1+y_3=0, z_1+z_3=2 n$ ...(iii)
Solving $\mathrm{eq}^{\mathrm{n}}$ (i), (ii) \& (iii), we get:
$\begin{aligned} & x_1=l, y_1=m, z_1=-n \\ & x_2=l, y_2=-m, z_2=n \\ & x_3=-l, y_3=m, z_3=n\end{aligned}$
So, the coordinate of $\mathrm{A} \equiv(l, m,-n), B \equiv(l,-m, n)$ and $C \equiv(-l, m, n)$
$\begin{aligned} & \therefore \frac{\mathrm{AB}^2+\mathrm{BC}^2+\mathrm{CA}^2}{l^2+m^2+n^2}=\frac{\left[\begin{array}{l}\left((0)^2+(2 m)^2+(2 n)^2\right) \\ +\left((2 l)^2+(2 m)^2+0^2\right) \\ +\left((2 l)^2+(2 n)^2\right)\end{array}\right]}{l^2+m^2+n^2} \\ & =\frac{4 m^2+4 n^2+4 l^2+4 m^2+4 l^2+4 n^2}{l^2+m^2+n^2} \\ & =\frac{8\left(l^2+m^2+n^2\right)}{\left(l^2+m^2+n^2\right)}=8\end{aligned}$