Equation of the Plane I have been working through all the problems in my textbook and I have finally got to a difficult one.
The problem ask 

Find the equation of the plane.The plane that passes through the points $(-1,2,1)$ and contains the line of intersection of the planes $x+y-z =2$ and $2x-y+3z=1$

So far I found the direction vector of the line of intersection to be $<1,-2,1>$ and I have identified a point on this line when $x=0$ to be $(0,5,-1)$. 
I do not know how to find the desired plane from here.
Any assistance would be appreciated.

Here is a related problem. To find the equation of the plane, you need three points which lie in the plane. Already, you have one, you can get the other two from the equation of the line, since the line lies in the plane. Once you have the three points, construct two vectors say $v_1$ and $v_2$ and find the cross product $n=v_1 \times v_2$. Then, the equation of the plane is given by
$$ n.((x,y,z)-(-1,2,1))=0 .$$
