What’s going on?
You can work everything out by breaking it down to the unit basis vectors, for which the products are explicit. One way of writing it, which feels more comfortable to me possibly at the expense of being the "wrong" sort of intuition, is
u(v^w) = (u.v)w - v(u.w) + u^v^w
(using . for inner product and ^ for outer product). I guess you can also think of the first two terms as the rule for taking the inner product of a vector and a wedge product, though I haven't thought through this completely.
Because of the repeated vector in the reflection formula, that last trivector vanishes, so the result is just a vector, as it will be whenever the vector is coplanar with the bivector.
I suppose this could be answered if you explicitly stated what set the geometric product acted on. As a guess, and from skimming Wikipedia, it’s the direct sum of scalars, vectors, bivectors, etc, up through n-vectors. So 2 + x∧y + 3x∧y∧z is a valid output. And it’s probably straightforward to show that the geometric product is actually defined on this space.
(Hi Marc!)
So, you basically have three case: x (xx) = x (1) = x -> a vector
x (xy) = x (x.y + x^y) = x (x.y) + xxy = x (x.y) + y -> a vector
x (yz) = x (y.z + x^y) = x (y.z) + xyz -> a vector + a trivector. This only happens if the three vectors are independent, which can never be the case for -ava