Solution :
$${\text{From the figure}}$$

$$\eqalign{
& {x_1} + {x_2} = 2l,\,{y_1} + {y_2} = 0,\,{z_1} + {z_2} = 0 \cr
& {x_2} + {x_3} = 0,\,{y_2} + {y_3} = 2m,\,{z_1} + {z_3} = 0 \cr
& {\text{and }}{x_1} + {x_3} = 0,\,{y_1} + {y_3} = 0,\,{z_1} + {z_3} = 2n \cr
& {\text{On solving, we get}} \cr
& {x_1} = l,\,{x_2} = l,\,{x_3} = - l \cr
& {y_1} = - m,\,{y_2} = m,\,{y_3} = m \cr
& {\text{and }}{z_1} = n,\,{z_2} = - n,\,{z_3} = n \cr
& \therefore \,{\text{Coordinates are}}\,: \cr
& A\left( {l,\, - m,\,n} \right),\,B\left( {l,\,m,\, - n} \right){\text{and }}C\left( { - l,\,m,\,n} \right) \cr
& \therefore \,\frac{{A{B^2} + B{C^2} + C{A^2}}}{{{l^2} + {m^2} + {n^2}}} \cr
& = \frac{{\left( {4{m^2} + 4{n^2}} \right) + \left( {4{l^2} + 4{n^2}} \right) + \left( {4{l^2} + 4{m^2}} \right)}}{{{l^2} + {m^2} + {n^2}}} \cr
& = 8 \cr} $$