= 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 $n\ne0$. A <ray> with nonzero direction $r$ approaches the front face only if
$$
n\cdot r<0.
$$
Reject $n\cdot r\geq0$. 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 is
$$
\lambda=\frac{n\cdot(a-o)}{n\cdot r}.
$$
Reject $\lambda<0$; 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 $u=b-a$, $v=c-a$, $w=p-a$, compute
$$
D=(u\cdot u)(v\cdot v)-(u\cdot v)^2>0,
$$
$$
\beta=\frac{(v\cdot v)(w\cdot u)-(u\cdot v)(w\cdot v)}D,\qquad
\gamma=\frac{(u\cdot u)(w\cdot v)-(u\cdot v)(w\cdot u)}D.
$$
These solve $p=a+\beta u+\gamma v$. The <barycentric coordinates> are $(1-\beta-\gamma,\beta,\gamma)$, so the complete acceptance criterion is
$$
\boxed{n\cdot r<0,\quad \lambda\geq0,\quad\beta\geq0,\quad\gamma\geq0,\quad\beta+\gamma\leq1;\qquad p=o+\lambda r.}
$$
The 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.
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