No, the 2d x-y plane in your example cannot pass through the 3d space’s origin because that would imply that you fixed the z coordinate at zero. The plane must be fixed at some nonzero z because you need to be able move x and y values by some scaled version of z to make translation happen. If z is zero, that scheme does not work.

Consider a transformation f where we wish to move x-y coordinates s units to the right. In 2d, we could express it as:

f(x, y) = (x + s, y)

But that transformation is affine not linear. There is no way to generate the value s as a linear combination of the inputs x and y. So, our workaround is to embed the x-y plane into 3d space at z=1. Then we can move (x,y,1) points in that plane s units to the right using this transformation:

f(x, y, z) = (x + s*z, y, z)

This new transformation is linear: it maps (0,0,0) to itself. But it maps our embedded 2d plane's origin (0,0,1) to (s,0,1), shifting it right by s units, as we want.

The matrix form of that transformation is:

    [[1 0 s]
     [0 1 0]
     [0 0 1]]

The same scheme would work if we had embedded the plane at any fixed z=r for nonzero r. We would only have to rescale the s in the matrix to s/r. Again, however, if r=0, this scheme will not work, as 1/r has gone to infinity.