The orthocentre of triangle formed by points: $(2,1,5)$, $(3,2,3)$ and $(4,0,4)$ is

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$A(2,1,5), B(3,2,3) 	ext { and } C(4,0,4)$ Equation of line $A C$ $\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z-5}{-1}$ Equation line $B C$ $\frac{x-3}{1}=\frac{y-2}{-2}=\frac{z-3}{1}$ Equation of altitude from $A$ on $B C$ $P$ is foot of perpendicular of $A P$ on $B C$ $\begin{aligned} & \Rightarrow P \equiv(r+3,-2 r+2, r+3) \ & 	ext {D.R.'s of } A P=(r+1,-2 r+1, r-2) \ & A P \perp B C \Rightarrow(r+1) 1+(-2 r+1)(-2)+(r+2) 1=0 \ & \Rightarrow r=\frac{1}{2} 	herefore P=\left(\frac{7}{2}, 1, \frac{7}{2}ight) \end{aligned}$ Equation of line $A P \frac{x-2}{-1}=\frac{y-1}{0}=\frac{z-5}{1}$ ....(iii) Similarly, let $Q$ be foot of perpendicular from $B$ to $A C$. $	herefore Q \equiv\left(3, \frac{1}{2}, \frac{9}{2}ight)$ Equation of line $B Q$ is $\frac{x-3}{0}=\frac{y-2}{3}=\frac{z-3}{-3}$ ....(iv) Solving (iii) and (iv) will give the orthocentre. From (iv), $\frac{x-3}{0}=\frac{y-2}{3}=\frac{z-3}{-3}=r$ $\Rightarrow(x, y, z)=(3,3 r+2,-3 r+3)$ Put in eq. (iii) $\frac{3-2}{-1}=\frac{3 r+2-1}{0}=\frac{-3 r+3-5}{1}$ $\Rightarrow r=-\frac{1}{3} \quad 	herefore$ Orthocentre is $(3,1,4)$.

The orthocentre of triangle formed by points: $(2,1,5)$, $(3,2,3)$ and $(4,0,4)$ is

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