Projection of an intersection of two planes on a third plane? I've been trying to solve this problem for about an hour now and I didn't manage to do it.
It goes like this:
I have two planes $x+y+z-2=0$, $x+2y+z-2=0$, they intersect. Find the projection of their intersection on a plane $3x+y+3z=1$.
Now, my plan was to form a line that two intersecting planes create. I did this by vector multiplication of two normal vectors of the two planes. 
I got a $vector(-1,0,1)$
Then by by solving the system of equations of two planes I got the general solution for a point that is within the line. 
$x+y+z=2$
$x+2y+z=2$
$y = 0$
$x = 2-z$
$z = z$, I then added a parameter $t$
$x = 2-t$
$y = 0$
$z = t$
And now I wanted to project two points from my line to the third plane. I did that by setting the value of $t$ to $0$, so for $t=0$
$p=\frac{x-2}{-1}=\frac{y-0}{0}=\frac{z-0}{2}=t$
so, 
$x = 2-t$
$y = 0$
$z = t$, and after that I inserted the values in the equation of the third plane so I got:
$3(2-t)+0+3t=1$
But as you can see here, the problem is that I get that
6 = 1, and t's are gone. 
Can you help me solve this, I would appreciate it very much, and thank you in advance.
 A: You made a good start by finding the line of intersection $L$ of the first two planes. Your next idea was also a good one. It looks you tried to find one of the projected points by computing the intersection of $L$ with the third plane, but there’s no such point: notice that $(-1,0,1)\cdot(3,1,3)=0$, so $L$ is parallel to the plane.  
This actually makes the problem a bit easier since you now know the direction of the projection is the same as that of $L$. All you need to do is to find one point on $L$’s projection, which you can do by projecting any point on $L$ onto the plane. This can be done in a variety of ways. For instance, taking $t=0$ as before, you have the point $(2,0,0)$ on $L$. The projection of this point onto the plane is then the point at which the perpendicular to the plane through this point intersects it. Using the plane’s normal from its equation, a parameterization of this perpendicular is $(2,0,0)+s(3,1,3)$. Substitute this into the plane’s equation and solve for $s$.  
This problem can also be solved without computing $L$ explicitly. Every plane that contains $L$ has an equation of the form $$(1-\lambda)(x+y+z-2)+\lambda(x+2y+z-2)=0.$$ The intersection of the plane onto which we’re projecting with the member of this family of planes that’s perpendicular to it is the projection of $L$. This perpendicularity condition leads to the equation $(1,\lambda+1,1)\cdot(3,1,3)=0$. Solve this for $\lambda$ to get the equation of the perpendicular plane through $L$ and then find the intersection of the two planes.
