How to calculate the intersection of two planes? How to calculate the intersection of two planes ?
These are the planes and the result is gonna be a line in $\Bbb R^3$:
$x + 2y + z - 1 = 0$
$2x + 3y - 2z + 2 = 0$

You need to solve the two equations
$$ x + 2y + z - 1 = 0 \\
2x + 3y - 2z + 2 = 0. $$
Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have
$$ x + 2y = 1-t \\
2x + 3y  = 2t-2. $$
Solving the last system gives 
$$ \left\{ x=-7+7\,t,y=4-4\,t \right\} .$$
Then the parametrized equation of the line is given by
$$ (x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t . $$
