The normal vectors to the two planes are (3, 3, 2) and (2, -3, 2). The cross product of these will be the direction vector of the line of intersection, (12, -2, -15).
Using x=0, we can find a point on this line by solving the simultaneous equations that remain:
... 3y +2z = -2
... -3y +2z = 2
Adding these, we get
... 4z = 0
... z = 0
so the point we're looking for is (x, y, z) = (0, -2/3, 0). This gives rise to the parametric equations ...
- x = 12t
- y = -2/3 -2t
- z = -15t
By letting t=2/3, we can find a point on the line that has integer coefficients. That will be (x, y, z) = (8, -2, -10).
Then our parametric equations can be written as
- x = 8 +12t
- y = -2 -2t
- z = -10 -15t
