Here is a geometrically flavored solution.
First test to see whether or not the triangles are parallel. If they are, then find the minimum distance $d$ between the two planes they occupy; if $d\ne0$ then the triangles do not intersect, if $d=0$ then our problem reduces to the planar case which we'll analyze shortly. For any $x\in\pi_1,y\in\pi_2$, two parallel planes $\pi_1$ and $\pi_2$ with shared normal vector $n$ have zero distance from each other if and only if $(x-y)\perp n$. The easiest selection would be to make $x,y$ vertices from distinct triangles.
In the planar case, two triangles intersect if and only if there is a pair of edges, one from each triangle, which intersect. Therefore the problem reduces to testing whether or not two line segments intersect (for nine possible pairs of segments, though only a maximum of seven need be tested for). This can be achieved with a basic least-squares method.
If the triangles are not parallel, then they intersect if and only if at least one edge of one of the triangles intersects the other triangle, and this intersection point (for a given edge) will be unique. Thus we will say the second triangle is affixed to the plane $\pi$, and because of symmetry only test whether a single edge from the first triangle intersects $\pi$, this edge having endpoints $a$ and $b$ (in reality, this test may need to be applied up to three times, once for each edge). Further, use an affine transformation on the entire setup so the second triangle has the origin $O$ as a vertex for convenience.
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Every point on the green line will be of the form $r\mathbf{a}+(1-r)\mathbf{b}, r\in[0,1]$. All points inside of the black triangle will be of the form $s\mathbf{u}+t\mathbf{v}, s,t\in[0,1]$. Equating the two expressions will have a unique solution as the point of intersection between the green line and the plane the black triangle lies on. Let $X$ denote the $3\times3$ matrix with columns $\mathbf{b}-\mathbf{a}$, $\mathbf{u}$, and $\mathbf{v}$. Then the point of intersection lies within the triangle if and only if $X^{-1} \mathbf{b} \in [0,1]^3$.