Find where the plane x-4y+3z=2 intersects the plane 9x+3y+z=5
1. Solve the system of simultaneous equations for x, y:
$\displaystyle \left|\begin{array}{l}x-4y=2-3z \\ 9x+3y=5-z\end{array}\right.$
2. You should come out with the result:
$\displaystyle \left|\begin{array}{l}x=\frac23-\frac13 z \\ y= -\frac13 + \frac23 z \\ z = z\end{array}\right.$
3. Replace z = t and you'll get the equation of the line as:
$\displaystyle \langle x,y,z \rangle = \langle \frac23, -\frac13 , 0\rangle + t \cdot \langle -\frac13, \frac23, 1 \rangle$