Yes. And I had an obvious typo "Such a vector is (2z, 2z, a)= z(2, 2, 1)" should have been "Such a vector is (2z, 2z, z)= z(2, 2, 1)" but you understood that.
Simlarly, if you have a subspace (x, y, z) satisfying only "y= z", x could be anything so such a vector would be (x, y, y)= (x, 0, 0)+ (0, y, y)= x(1, 0, 0)+ y(0, 1, 1). {(1, 0, 0), (0, 1, 1)} would be a basis and the dimension would be 2.