Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. Let $A D, B E$ be altitudes. Then $B D=5, C D=9$. (If you don't already know this, it can be deduced from the Pythagorean Theorem: $C D^{2}-B D^{2}=\left(C D^{2}+A D^{2}\right)-\left(B D^{2}+A D^{2}\right)=A C^{2}-A B^{2}=56$, while $C D+B D=B C=14$, giving $C D-B D=56 / 14=4$, and now solve the linear system.) Also, $A D=\sqrt{A B^{2}-B D^{2}}=12$. Similar reasoning gives $A E=33 / 5$, $E C=42 / 5$. Now let $F$ be the point on $B C$ such that $C F=B D=5$, and let $G$ be on $A C$ such that $C G=A E=33 / 5$. Imagine placing face $A B C$ flat on the table, and letting $X$ be a point in space with $C X=13, B X=14$. By mentally rotating triangle $B C X$ about line $B C$, we can see that $X$ lies on the plane perpendicular to $B C$ through $F$. In particular, this holds if $X$ is the fourth vertex of our tetrahedron $A B C X$. Similarly, $X$ lies on the plane perpendicular to $A C$ through $G$. Let the mutual intersection of these two planes and plane $A B C$ be $H$. Then $X H$ is the altitude of the tetrahedron. To find $X H$, extend $F H$ to meet $A C$ at $I$. Then $\triangle C F I \sim \triangle C D A$, a 3-4-5 triangle, so $F I=C F \cdot 4 / 3=20 / 3$, and $C I=C F \cdot 5 / 3=25 / 3$. Then $I G=C I-C G=26 / 15$, and $H I=I G \cdot 5 / 4=13 / 6$. This leads to $H F=F I-H I=9 / 2$, and finally $X H=\sqrt{X F^{2}-H F^{2}}=\sqrt{A D^{2}-H F^{2}}=3 \sqrt{55} / 2$. Now $X A B C$ is a tetrahedron whose base $\triangle A B C$ has area $A D \cdot B C / 2=12 \cdot 14 / 2=84$, and whose height $X H$ is $3 \sqrt{55} / 2$, so its volume is $(84)(3 \sqrt{55} / 2) / 3=42 \sqrt{55}$.