Find the equation of the plane that goes through (0,0,2) and the line y = x I'm trying to find the equation of the plane that goes through $(0,0,2)$ and the line $ y = x$.   So far, I attempted taking the points $ P = (1,1,0), Q = (2,2,0), R = (0,0,2)$ and find $PQ,PR$ make the cross product $n$ and find the solution of $n*(z-P) = 0$, where $z = (x,y,z)$ is an arbitrary point in the plane, but after computing I get $x+y = 2$ which is a line and not a plane, I cannot see my mistake, please could someone point me in the right direction?.  Thanks !
Hint:

In $\mathbb{R}^3$ the equation $y=x$ does not represents a line but represents the plane that contain the $z$ axis and the line in the $xy-$plane that bisects the angle between the $x$ and the $y$ axis. So this plane contains the point $(0,0,2)$.  

I think that your mistake is in the calculus of the vector $\vec n$ that is $\vec n =(2,-2,0)^T$. So that your $\vec n \cdot (\vec z- P)=0$ becomes $2x-2y=0$, that is $y=x$.
I didn't fact-check your calculus, but $x+y = 2$ is not a line in $\mathbb{R}^3$, but indeed a plane.

You have to imagine it like this: in $\mathbb{R}^2$, $x+y =2$ describes a line. But in $\mathbb{R}^3$, since $z$ can be everything in $\mathbb{R}$, all the points satisfying $x+y = 2$ in $\mathbb{R}^3$ are on a line for a fixed $z$, but by varying $z$ through $\mathbb{R}$, you move the line along the $z$-axis, and you get a plane.

The set of points you have is:

$$\{(s,2-s,t) \in \mathbb{R}^3 \mid s,t \in \mathbb{R}\}$$

That's a $2$-dimensional affine space in $\mathbb{R}^3$, so it's a plane.

EDIT: as pointed out by Doug M in the comments, the line $y = x$ does not lie in the set that I mentioned. So the calculus is indeed incorrect. Note that $y=x$ is not a line either, but a plane, for the same argument as above. And $(0,0,2)$ belongs to that plane, so it's redundant information. Maybe there's some mistake in the question.