Let $b_1,b_2,b_3$ be the normalized normals to the planes and $\s_j$ the reflection about the plane normal to $b_j$. Then $b_1,b_2,b_3$ forms an orthonormal basis, $\s_j(b_j)=-b_j$ and for $j\neq k$: $\s_j(b_k)=b_k$. It folows that $\s_1\s_2\s_3(b_j)=-b_j$, i.e. $\s_1\s_2\s_3$ equals the inversion.