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

The midpoints of the sides are given as,
$AB,BC$ and $CA$ are$\left(3,0,0\right),\left(0,4,0\right),\left(0,0,5\right)$

The coordinates of $A,B$ and $C$
$A=\left(3,-4,5\right)$
$B=\left(3,4,-5\right)$
$C=\left(-3,4,5\right)$

And
$A{B}^2={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2$

$=(3-3{)}^2+(4+4{)}^2+(-5-5{)}^2$

$=8^2+{10}^2$

$=164$
Similarly,
$B{C}^2=(-3-3{)}^2+(4-4{)}^2+(5+5{)}^2$

$=36+100$

$=136$

Similarly,

$C{A}^2=(-3-3{)}^2+(4-4{)}^2+(5+5{)}^2$

$=36+64$

$=100$

$A{B}^2+B{C}^2+C{D}^2=164+136+100$

$=400$