Suppose we are given a plane in $\mathbb{R}^{3}$ with its scalar equation, we are also given two vectors $\vec{x},\vec{y}$, is it possible to determine which of $\vec{x},\vec{y}$ is closer to the plane $P$ by only calculating $proj_{P}\vec{x}$ and $proj_{P}\vec{y}$?
Edit: To be clear, let's say $P:3x_1+x_2-2x_3=0, \vec{x}=(-1,2,1),\vec{y}=(1,0,2)$, is it possible to determine which of $\vec{x},\vec{y}$ is closer to $P$ by only calculating the projections?
Possible with only the projected vectors? No.
Consider the XY plane (z=0). And two vectors, (x,y,z1), (x,y,z2) (notice the same x,y). If you do an orthogonal projection to the XY plane, boths projections are the same, the z info is lost.
Generally speaking, all vectors in the same romboid defined by a vector and a direction of projection have the same projection. For a perpespective projection, given a vector you can find another vector (with other size) contained in the plane of projection that has the same projection.