$\left(0,4,0\right)$In a triangle ABC, if the mid points of sides AB, BC, CA are $\left(3,0,0\right),\left(0,4,0\right),\left(0,0,5\right)$ respectively, then ${\mathrm{AB}}^2+{\mathrm{BC}}^2+{\mathrm{CA}}^2=$

Let $A\left(x_1,y_1,z_1\right),B\left(x_2,y_2,z_2\right),C\left(x_3,y_3,z_3\right)$ are vertices of a triangle

Given $\left(3,0,0\right)$ is mid point of $AB$, so that $x_1+x_2=6,y_1=-y_2,z_1=-z_2$

Given $\left(0,4,0\right)$ is midpoint of $BC$, so that $x_2+x_3=0,y_2+y_3=8,z_2+z_3=0$

Given $\left(0,0,5\right)$ is midpoint of $AC$, so that $x_1+x_3=0,y_1+y_3=0,z_1+z_3=10$

it implies that $A\left(3,-4,5\right),B\left(3,4,-5\right),C\left(-3,4,5\right)$

Hence, 

  $\begin{array}{rcl}{\left(AB\right)}^2+{\left(BC\right)}^2+{\left(CA\right)}^2& =& 64+100+36+100+36+64\\ & =& 400\end{array}$