3D intersection of a line segment with a triangle via convexity I thought about detecting whether a line segment intersects a triangle and came up with the idea of using convexity, namely that the shape one gets from spanning faces from the line segment start point to the triangle to the line segment end point is a convex polyhedron iff the line intersects.
(The original triangle is not a face of that shape!)

How could this be proved or refuted?
 A: First, let me address a slightly different version of your conjecture, where you take the union of the polyhedra spanned by each point and the triangle.
Let me fix some notation.  Let us say we are considering the line segment $S$ from a point $p$ to a point $q$, and asking whether it intersects a triangle $T$.  The conjecture is then that $S\cap T\neq \emptyset$ iff $X=\operatorname{conv}(p\cup T)\cup \operatorname{conv}(q\cup T)$ is convex, where $\operatorname{conv}$ denotes convex hull.  The forward direction of this conjecture is true, but the reverse direction is not true.  In fact, more generally the following is true:

Theorem: Let $V$ be a real vector space, $T\subseteq V$ be a convex set, and $p,q\in V$ be two distinct points.  Let $L$ be the line (not line segment!) between $p$ and $q$.  Then $L\cap T\neq \emptyset$ iff $X=\operatorname{conv}(p\cup T)\cup \operatorname{conv}(q\cup T)$ is convex.

In particular, if the line between $p$ and $q$ intersects $T$ but the line segment between them does not, this gives a counterexample to the conjecture.
Let us now prove this theorem.  First we prove the forward direction.  Suppose that $L\cap T\neq \emptyset$; let $r\in L\cap T$.  To show that $X$ is convex, by convexity of $T$ it suffices to show that $ap+bq+ct\in X$ whenever $t\in T$ and $a,b,c\geq 0$, $a+b+c=1$.  If $a=b=0$ this is trivial, so we may assume $a+b>0$.  Note that $\frac{ap+bq}{a+b}\in L$ and is between $p$ and $q$.  It follows that either $\frac{ap+bq}{a+b}$ is between $p$ and $r$ or $\frac{ap+bq}{a+b}$ is between $q$ and $r$.  Let us assume WLOG that $\frac{ap+bq}{a+b}$ is between $p$ and $r$.  Since $r\in T$, this means that $\frac{ap+bq}{a+b}$ is in $\operatorname{conv}(p\cup T)$.  But $ap+bq+ct=(a+b)\frac{ap+bq}{a+b}+ct$ is on the line segment between $\frac{ap+bq}{a+b}$ and $t$, and hence is also in $\operatorname{conv}(p\cup T)$.  Thus $ab+bq+ct\in X$, as desired.
Now for the reverse direction.  Suppose $L\cap T=\emptyset$.  Note that any point of $\operatorname{conv}(p\cup T)$ is on a line segment from $p$ to a point of $T$, by convexity of $T$.  Since $L\cap T=\emptyset$, such a line segment is not a segment of $L$, so it intersects $L$ at at most one point, namely $p$.  Thus $\operatorname{conv}(p\cup T)\cap L=\{p\}$.  Similarly, $\operatorname{conv}(q\cup T)\cap L=\{q\}$.  But since $p\neq q$, this means that $X$ cannot contain the entire line segment between $p$ and $q$.  Thus $X$ is not convex.

Let me now relate the Theorem above to your intended statement.  Let me assume that $p$ and $q$ are not in the plane of $T$ (if they are, I don't know how you intend to define your polyhedron).  Note that if $S\cap T\neq \emptyset$, then $p$ and $q$ must be on opposite sides of the plane of $T$, and so your polyhedron is the same as my $X$.  Thus my proof above still shows that if $S\cap T\neq\emptyset$, then the polyhedron is convex.
Conversely, note that your polyhedron is always contained in my $X$, and also contains $p$, $q$, and the vertices of $T$.  Thus if your polyhedron is convex, so is $X$ (since they must both then be the entire convex hull of $p$, $q$, and the vertices of $T$).  So in that case $L\cap T\neq \emptyset$.  If $L\cap T\neq\emptyset$ but $S\cap T=\emptyset$, that means that $p$ and $q$ are on the same side of the plane of $T$; WLOG assume that the point of $L\cap T$ is closer to $q$ than to $p$.  In that case your polyhedron will just be the tetrahedron spanned by $p$ and $T$ with the tetrahedron spanned by $q$ and $T$ removed.  This polyhedron will not be convex (since, for instance, it does not contain the interior of $T$).
Thus the only way your polyhedron can be convex is if $L\cap T\neq\emptyset$ and the intersection point is actually in $S$.  This proves your conjecture.
