The answer is going to be in two parts.

Answer, part 1:

First we need to understand what we really mean when we say "a point lies on a plane." Recall that by convention we represent a plane P by the row vector [a b c d] and a point V by the column vector [x y z 1]^T.

Now consider what happens when we transform all the vertices of a shape by some matrix M. In that case, every vertex V is replaced by V' = (M • V).

So how then would we transform planes? Clearly we want to transform planes so that they still contain the same vertices. In other words, we want to find P' such that:

But this means that we want P' = (P • M^-1), since then:

P' • V' =
(P • M^-1) • (M • V) =
P • (M^-1 • M) • V =
P • V = 0

Answer, part 2:

Now consider the surface normal direction vector N = [n_x n_y n_z 0] at some vertex V. The three values n_x,n_y,n_z are really describing the three linear components (the "a b c" parts) of the tangent plane P = [a b c d] which passes through the surface at vertex V. Transforming N is simply a matter of transforming that plane P, and then throwing out its "d" coefficient. In other words, want N' = N • M^-1.

And now we've answered the original question, since: