Is what I'm doing valid? 
Find the POI of the following two planes:
$$\pi_1: -3x + 3y  + z + 6= 0$$
$$\pi_2: 3x - y + 2z - 2 = 0$$

I started by isolating "$z$".

$$\pi_1: z = 3x - 3y - 6$$
$$\pi_2: z = \frac {-3x + y + 2}{2}$$

I then made them equal to each other: $z_1 = z_2$.
Therefore

$3x - 3y - 6 = \frac {-3x + y + 2}{2}$

Moving things around the above equation is $9x - 7y - 15 = 0$. Is this the equation of the line of intersection? I'm pretty sure my method is wrong because using my method you always get a line lying on a plane (i.e. the above line lies on the XY plane) because you always get rid of one variable.
If my method is wrong, how do I correctly solve the question, without changing forms and just solving the system of equations?
 A: Two planes can be parallel (you can know this by looking at the normal vector of the two planes: if the the normal is parallel, the corresponding planes are.) If they are not parallel, they must intersect at infinitely many points, i.e. a line of intersection must exist between two planes, which is exactly what has happened in your case. 
Another way to think about this is: you have two equations and three independent variables which clearly tells that a line of intersection defined by the cross product of two normal vectors must exist. 
A: Hint:The equation for the plane of intersection of planes $\pi_1,\pi_2$ is $\pi_1-\lambda\pi_2=0$ for some $\lambda \in R$
A: The normal vector for the first plane equals $(-3, 3, 1)$ and the normal vector to the second plane is $(3, -1, 2)$. Take the cross product of them, this is $v = (7,9,-6)$ (if I made no error). This vector is prependicular to both normal vector and so must be a direction vector that lies in both planes, hence the direction of the line of intersection. 
Now find (using linear algebra) any one point $p$ in the intersection.
Then $p + tv$ must be the line of intersection.
