Does the sphere $x^2 + (y-2)^2 + z^2 = 1$ intersect the plane $z-x = 3$? If so, find a point in their intersection. If not explain why. 
Does the sphere $x^2 + (y-2)^2 + z^2 = 1$ intersect the plane $z-x = 3$? If so, find a point in their intersection. If not explain why.

I feel like I can give a convincing answer with a rough sketch, and words, but I can not figure out how to solve the problem algebraically. If anyone could help I would be greatly appreciative!

If so, $$x^2+(y-2)^2+(x+3)^2=1$$ or
$$2(x^2+3x+4)+(y-2)^2=0,$$ which is impossible.
