= Solution
Let the <triangles> have vertices $A_0,A_1,A_2$ and $B_0,B_1,B_2$. Work initially with nonzero-area <triangles>. A <triangle-triangle intersection algorithm> begins with the unnormalized <normal vector> $N_A=(A_1-A_0)\times(A_2-A_0)$ and the three signed plane evaluations $d_j=N_A\cdot(B_j-A_0)$. If all $d_j$ are strictly positive, or all strictly negative, the second <triangle> lies strictly on one side of the first one's plane, so the <triangles> are disjoint. Otherwise compute $N_B$ and $e_i=N_B\cdot(A_i-B_0)$ and apply the symmetric rejection test. Strict inequalities are important: contact on a vertex or edge counts as intersection.
If the planes are not parallel, put $L=N_A\times N_B$. Each <triangle> meets the other one's plane in an empty set, a point or a <line segment>. Its endpoints are obtained from zero-valued vertices and the edges whose endpoint plane evaluations have opposite signs. For example, an edge $B_jB_k$ contributes
$$
X=B_j+\frac{d_j}{d_j-d_k}(B_k-B_j).
$$
All these points lie on the common line of the two planes. Choose a coordinate $r$ for which $L_r\ne0$ and form the coordinate intervals $I_A=[a_-,a_+]$, $I_B=[b_-,b_+]$ of the two slices. This coordinate is one-to-one along the common line, so the exact final test is
$$
\boxed{\max(a_-,b_-)\le\min(a_+,b_+).}
$$
Only that coordinate of each slice endpoint needs to be evaluated; no full three-dimensional line parameterization is necessary.
Parallel planes which have survived the strict plane tests are coplanar. Drop the coordinate corresponding to the largest component of $N_A$ and test the two projected planar <triangles>. For each edge of either projected <triangle>, project both vertex sets onto the perpendicular direction. A strictly disjoint pair of projection intervals certifies separation. If none of the six edge directions separates them, the planar <convex sets> overlap: a separating line for two disjoint convex polygons can be chosen parallel to an edge of one of them. This includes containment and boundary contact.
A zero <normal vector> signifies a degenerate <triangle>; reduce its <convex hull> to a point or its extreme <line segment> and use point containment or segment intersection instead. For floating-point input, reliable sign predicates and scale-aware error bounds are preferable to treating a tiny signed distance as an arbitrary exact zero. \b[Plane rejection followed by interval overlap, with a planar test for coplanar input, determines all nondegenerate cases.]
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