The given planes are $2x – y + z = 1$ and $3x + y + z = 2$.
Adding the two equations I choose the two points of intersection of the planes to be $P= (3, -1, -6)$ and $X = (5, -2, -11)$. Then, the equation of the line is $(5, -2, -11)= (3, -1, -6) + tA$. So the vector $tA = (2, -1, -5)$. Any vector parallel to $tA$ must be parallel to the line. Let $t = -1$. Then $A = (-2, 1, 5)$ is parallel to $tA$.
I am not sure if my solution is correct. I just noticed that the problem that comes right after this one asks to find a parametric representation for the line of intersection of the planes of the given equations above. It suggests that they might have solved the problem in a different manner. I looked up the answer to that question and it's $X = (1, 0, -1) + t(-2, 1, 5)$. My vector A and theirs don't match. So I might have made a mistake.
Please, elaborate. Thanks.
edit: Turns out vectors do match!
Best Answer
The cleanest way to do this uses the vector product: if $\mathbf{n_1}$ and $\mathbf{n_2}$ are the normals to the planes, then the line of intersection is parallel to $\mathbf{n_1} \times \mathbf{n_2}$.
The equation of a plane takes the form $\mathbf{x.n} = a$, where $\mathbf{x} = (x,y,z)$ and $a$ is a scalar constant. So in your case we have:
$\mathbf{n_1} = (2,-1,1)$ (from $2x−y+z=1$)
$\mathbf{n_2} = (3,1,1)$ (from $3x+y+z=2$)
So the line of intersection is parallel to $\mathbf{n_1} \times \mathbf{n_2} = (-2,1,5)$.