Solution:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $AD$, $BE$ be altitudes. Then $BD = 5$, $CD = 9$. (If you don't already know this, it can be deduced from the Pythagorean Theorem: $CD^{2} - BD^{2} = (CD^{2} + AD^{2}) - (BD^{2} + AD^{2}) = AC^{2} - AB^{2} = 56$, while $CD + BD = BC = 14$, giving $CD - BD = 56 / 14 = 4$, and now solve the linear system.) Also, $AD = \sqrt{AB^{2} - BD^{2}} = 12$. Similar reasoning gives $AE = 33 / 5$, $EC = 42 / 5$.

Now let $F$ be the point on $BC$ such that $CF = BD = 5$, and let $G$ be on $AC$ such that $CG = AE = 33 / 5$. Imagine placing face $ABC$ flat on the table, and letting $X$ be a point in space with $CX = 13$, $BX = 14$. By mentally rotating triangle $BCX$ about line $BC$, we can see that $X$ lies on the plane perpendicular to $BC$ through $F$. In particular, this holds if $X$ is the fourth vertex of our tetrahedron $ABCX$. Similarly, $X$ lies on the plane perpendicular to $AC$ through $G$. Let the mutual intersection of these two planes and plane $ABC$ be $H$. Then $XH$ is the altitude of the tetrahedron.

To find $XH$, extend $FH$ to meet $AC$ at $I$. Then $\triangle CFI \sim \triangle CDA$, a 3-4-5 triangle, so $FI = CF \cdot 4 / 3 = 20 / 3$, and $CI = CF \cdot 5 / 3 = 25 / 3$. Then $IG = CI - CG = 26 / 15$, and $HI = IG \cdot 5 / 4 = 13 / 6$. This leads to $HF = FI - HI = 9 / 2$, and finally $XH = \sqrt{XF^{2} - HF^{2}} = \sqrt{AD^{2} - HF^{2}} = 3 \sqrt{55} / 2$.

Now $XABC$ is a tetrahedron whose base $\triangle ABC$ has area $AD \cdot BC / 2 = 12 \cdot 14 / 2 = 84$, and whose height $XH$ is $3 \sqrt{55} / 2$, so its volume is $(84)(3 \sqrt{55} / 2) / 3 = 42 \sqrt{55}$.