The orthocenter of triangle formed by the points (2, 1, 5)  (3, 2,3) and (4,0, 4)  is

Let  A(  2,1,5) B ( 3,2,3) , C ( 4, 0, 4)

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

$\mathrm{BC}=\sqrt{(3-4{)}^2+(0-2{)}^2+(4-3{)}^2}=\sqrt{6}$

$\mathrm{CA}=\sqrt{(4-2{)}^2+(0-1{)}^2+(4-5{)}^2}=\sqrt{6}$

$∆\mathrm{ABC}$ is an equilateral

Orthocenter $∆\mathrm{ABC}$=centroid $∆\mathrm{ABC}$=  ( 3, 1, 4)