If Vertices of a $\triangle ABC$ are $P(p,0,0),Q(0,q,0),R(0,0,r),$ Then orthocenter of a $\triangle ABC$ is
$\bf{My\; Try::}$ Point of intersection of Altitudes is called orthocenter.
Now equation of plane which passes through $P,Q,R$ is $\displaystyle \frac{x}{p}+\frac{y}{q}+\frac{z}{r} = 1$
Where $\displaystyle \left<\frac{1}{p},\frac{1}{q},\frac{1}{r}\right>$ be the Direction ratio of a line Which is perpendicular to the plane.
Now How can i solve it , Help required, Thanks
Pick an arbitrary point $X=(x,y,z)$ on the altitude drawn from $P$ onto $QR$. Now, $$PX\perp QR\implies \vec{PX}\cdot\vec{QR}=0\implies yq=zr$$ Also $X$ satisfies the equation of the plane of the triangle. Using these facts we can write the parametric equation of the altitude drawn from $P$ as $$ x=p\left(1-\frac{tr}{q}-\frac{tq}{r}\right), y=rt, z=qt $$ Similarly, altitude drawn from $Q$ has the parametric equation $$ x=rk, y=q\left(1-\frac{kp}{r}-\frac{kr}{p}\right), z=pk $$ Solving these equations we get $$ (x,y,z)=\frac{p^2q^2r^2}{p^2q^2+q^2r^2+r^2p^2}\left(\frac{1}{p},\frac{1}{q},\frac{1}{r}\right),\qquad \left(k=\frac{q}{p}t,t=\frac{p^2qr}{p^2q^2+q^2r^2+r^2p^2}\right) $$ which gives the orthocenter.