In a $\mathrm{△}ABC$  the mid-point of the sides $AB, BC$and $CA$ are respectively$(l,0,0), (0,m,0) \text{ and }(0,0,n)$ respectively

Then, $\frac{A{B}^2+B{C}^2+C{A}^2}{l^2+m^2+n^2}=$

We know that in a $\mathrm{△}ABC$  if the coordinates of the mid-

points of sides $BC, CA and AB$ are respectively $D\left(x_1,y_1,z_1\right)$

$E\left(x_2,y_2,z_2\right)$ and $F\left(x_3,y_3,z_3\right)$ then the coordinates of its vertices 

are $A\left(-x_1+x_2+x_3,-y_1+y_2+y_3,-z_1+z_2+z_3\right)$

$\begin{array}{l}B \left(x_1-x_2+x_3,y_1-y_2+y_3,z_1-z_2+z_3\right)\text{ and }\\ C \left(x_1+x_2-x_3,y_1+y_2-y_3,z_1+z_2-z_3\right).\end{array}$

Thus, the coordinates of $A , B ,C$ are $(l,-m,n), B(l,m,-n)$

and  $C (-l,m,n)$ 

$\therefore \frac{A{B}^2+B{C}^2+C{A}^2}{l^2+m^2+n^2}=\frac{4\left(m^2+n^2\right)+4\left(l^2+n^2\right)+4\left(l^2+m^2\right)}{l^2+m^2+n^2}=8$