Determine the vector function r(y) whose graph is the cross-section of the graph of z = f(x,y) = x^2 - 2y^2 + x - 3y by the plane x = 2.

I was attempting to do this one alone, but I am confused as to how a vector is created from only functions that include x and y and no z?

I am leaning towards the answer of r(y) = < 2, y, 6 - 2y^2 - 3y >, solely from general knowledge and plugging x=2 into z. Therefore, z turns into a function of y, x is already 2, and y is just... y.

Is this correct?