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.

To find a plane in space, you need  a normal direction vector (A,B,C) and a point. The equation is $A(x-x_0) + B(y-y_0) + C(z-z_0)=0$. In your case, that is
$A(x+1)+B(y-2)+C(z-1)=0.$ Now take the cross product of the normals to your other planes $(1,1,-1)\times(2,-1,3)$ to get a vector that is perpendicular to the line of intersection. 
