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

$\sqrt{3}$

Let P(1, -2, 3) and R(2, -3, 5) be the coordinates of the extremities P and R of a diagonal of the square PQRS. Then,

$PR= \sqrt{(2-1)^2+(-3+2)^2+(5-3)^2}= \sqrt{6}$

∴ Length of the side $= \frac{\sqrt{6}}{\sqrt{2}}= \sqrt{3}$        $[∵ Diagonal = \sqrt{2}(side)]$