find coodinates of orthocentre in 3d plane The plane $$2x+y+2z=9$$ intersects te coordinate axes in $A$, $B$, $C$ now we have to find the coordinates of the orthocenter of the triangle $ABC$.
I tried and get to know that the coordinates of triangle above are:
$$\left(\frac92,0,0\right),$$
$$(0,9,0),$$
$$\left(0,0,\frac92\right).$$
But now how to proceed? 
 A: This is what we have so far

First write up the equation of the straight line through $C$ in the direction of the vector $b=\left[\begin{smallmatrix}\ \ 0\\-9\\\ \ \frac92\end{smallmatrix}\right]$ pointing from $B$ to $C$. Here it is
$$r(t)=\begin{bmatrix}\ 0\\-9\\\ \frac92\end{bmatrix}t+\begin{bmatrix}0\\0\\\frac92\end{bmatrix}.$$
Then take a vector pointing to an arbitrary  point of this straight line from $A$:
$$v(t)=\begin{bmatrix}\frac92\\9t\\-\frac92t-\frac92\ \end{bmatrix}$$
Then take the scalar product of $v$ and $b$:
$$v(t)\cdot b=-t\left(81+\frac{81}4\right)-\frac{81}4.$$
If we want $v$ to be perpendicular to $b$ then we have to solve the equation $$b\cdot v(t)=0$$ for $t$. The result is $t=-\frac15.$ the vector perpendicular to $b$ is 
$$v\left(-\frac15\right)=\begin{bmatrix}\ \ \ \frac92\\-\frac95\\-\frac9{10}\end{bmatrix}.$$
Then, the equation of the straight line through $A$ in the direction of this $v$ is
$$u(t)=\begin{bmatrix}\ \ \ \frac92\\-\frac95\\-\frac9{10}\end{bmatrix}t+\begin{bmatrix}\frac92\\0\\0\end{bmatrix}.$$
The same way, we can determine the straight line $w(t)$ trough $B$ in the direction of the vector from $B$ to $M$ (the mid point of $CA$). The intersection point of $u(t)$ and $w(t)$ the the two straight line will be the orthocenter of the triangle $ABC$.
Double check my calculations!
A: Here's another way to solve it :
We will solve a generalized version of this problem for any plane not passing through the origin, intersecting the $x,y,z$ - axis at points $A,B,C$ respectively.
We claim the orthocenter of triangle $ABC$ is the projection of the origin $O$ on the triangle, say $H$.
Observe that $AO$ is perpendicular to plane $ OBC $, so $AO \perp BC$.
Also $OH$ is perpendicular to plane $ABC$, so $OH \perp BC$.Hence the plane $AOH$ is perpendicular to $BC$, so $AH \perp BC$. Similarly $BH\perp CA$ and $CH \perp AB$. Hence $H$ must be the orthocenter of triangle $ABC$ as desired. $\blacksquare$
