Find a vector parallel to the line of intersection of the two given planes. 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!  
 A: 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 $\mathbf{n_1} \times \mathbf{n_2} = (-2,1,5)$.
A: An alternative approach (in case you are not creative enough to find two such points)
Let $z=t$. Then we have that $2x-y = 1-t$ and $3x + y = 2-t$. Now, we solve and get $x = 1/5[1\cdot (1-t) + 1\cdot (2-t)] = 3/5 - 2t/5$ and $y=1/5[-3\cdot (1-t) + 2\cdot (2-t)]= 1/5+t/5$. So $(x,y,z) = (3/5,1/5,0) + t(-2/5, 1/5, 1)$. This is equivalent to $(x,y,z) = (3/5, 1/5, 0) + t(-2, 1, 5)$, since we are either stretching or dilating a line. Now, $(3/5,1/5,0) + (-1/5)\cdot(-2,1,5) = (1, 0, -1)$ and so this point is on the line. Indeed, we can write $(x,y,z) = (1,0,-1) + t(-2, 1, 5)$ as desired.
Your answer is fine, it just shifts the t-value
As for your answer, the problem is fine. Find two points of intersection and take their difference. So, we have $(x,y,z) = (3,-1,-6) + (u-1)(-2,1,5) + u(-2,1,5) = (1, 0, -1) + t(-2,1,5)$ where $t = u-1$ and $u,t \in \mathbb{R}$ as desired.
If guessing, a common approach is to set let's say $z=0$ in one case and $z=1$ in another and solve two simultaneous equations to find the points. Maybe you guessed your points?
