Let $A =(1,2,3) $
$B =(3,-1,5) $
$C =(4,0,-3)$
$D R's$ of $\overline{A B}=(2,-3,2)$
$D R'$ s of $\overline{B C}=(1,1,-8)$
$D R's $ of $\overline{A C}=(3,-2,-6)$
Here, we noticed that $ \overline{A B} \perp \overline{A C}$
$\Rightarrow \, \angle A=90^{\circ}$
Orthocentre of $\Delta\, A B C $ = vertex A
$H =(1,2,3)$
Circumcentre of $\Delta \,A B C=$ mid-point of $\overline{B C}$
$S=\left(\frac{7}{2}, \frac{-1}{2}, 1\right)$
$\therefore $ Required distance $(H S)=\sqrt{\frac{33}{2}}$
$\sqrt{\left(\frac{7}{2}-1\right)^{2}+\left(-\frac{1}{2}-2\right)^{2}+(1-3)^{2}} $
$=\sqrt{\frac{25}{4}+\frac{25}{4}+4}=\sqrt{\frac{66}{4}}$
