If the extremities of a diagonal of a square are (1,2,3) and (2,–3,5), then the its side is of length

The ends of diagonals of a square are $A\left(1,2,3\right),B\left(2,-3,5\right)$

Suppose that the length of side of the square is $x$

$x^2+x^2={\left(AB\right)}^2$

It implies that 

$\begin{array}{rcl}{x}^2& =& \frac{1}{2}\left({\left(2-1\right)}^2+{\left(-3-2\right)}^2+{\left(5-3\right)}^2\right)\\ & =& \frac{1}{2}\left(1+25+4\right)\\ & =& \frac{30}{2}\\ & =& 15\end{array}$

Therefore, the length of the side of the square is $x=\boxed{\sqrt{\mathbf{15}}}$