 Solution

Let the oriented vertices be $a,b,c$, in an order such that $n=(b-a)\times (c-a)$ points out of the front face. The given nondegeneracy guarantees $n\ne 0$. A ray with nonzero direction $r$ approaches the front face only ifReject $n\cdot r\ge 0$. This combines back-face culling with rejection of parallel rays, including the coplanar case for this ordinary surface-hit test. The ray-plane intersection parameter isReject $\lambda <0$; for rays launched from existing surfaces use a positive lower cutoff to avoid counting the launch surface. Form $p=o+\lambda r$.

Finally determine whether $p$ lies inside the triangle. For $u=b-a$, $v=c-a$, $w=p-a$, computeThese solve $p=a+\beta u+\gamma v$. The barycentric coordinates are $(1-\beta -\gamma ,\beta ,\gamma )$, so the complete acceptance criterion isThe inequalities include the edges and vertices. A renderer sharing edges between triangles should use a consistent numerical edge convention; small floating-point arithmetic errors near a shared boundary should not create cracks. Normalizing $n$ is unnecessary for this intersection calculation because its scale cancels.