[ Sphere inscribed in a pyramid ]
[Volume helps solve the problem.]

In a triangular pyramid, two opposite edges are equal to 12 and 4, and the other edges are equal to 7. A sphere is inscribed in the pyramid. Find the distance from the center of the sphere to the edge equal to 12.
Let $A B C D$ be a triangular pyramid, where $A B=12, C D=4, A C=B C=A D=B D=7$. Since $D A = D B$, the orthogonal projection $P$ of vertex $D$ onto the plane $A B C$ is equidistant from points $A$ and $B$, so point $P$ lies on the perpendicular bisector of side $A B$ of the isosceles triangle $A B C$. Let $K$ be the midpoint of $A B$. Then

$$
D K=C K=\sqrt{A C^{2}-A K^{2}}=\sqrt{49-36}=\sqrt{13}
$$

If $M$ is the midpoint of side $C D$ of the isosceles triangle $C D K$, then $K M$ is the height of this triangle, so

$$
K M=\sqrt{C K^{2}-C M^{2}}=\sqrt{13-4}=3
$$

and since $D P$ is also the height of triangle $C K D$, then

$$
D P=\frac{C D \cdot K M}{C K}=\frac{4 \cdot 3}{\sqrt{13}}=\frac{12}{\sqrt{13}}, \sin \angle C K D=\frac{D P^{P}}{D K}=\frac{\frac{12}{\sqrt{13}}}{\sqrt{13}}=\frac{12}{13}
$$

Let $r$ be the radius of the sphere inscribed in the pyramid $A B C D, S$ be the total surface area of the pyramid, and $V$ be its volume. Then

$$
\begin{gathered}
V=\frac{1}{3} \cdot \frac{1}{2} A B \cdot C K \cdot D P=\frac{1}{3} \cdot \frac{1}{2} \cdot 12 \sqrt{13} \cdot \frac{12}{\sqrt{13}}=24 \\
S_{\triangle B C D}=S_{\triangle A C D}=\frac{1}{2} C D \cdot \sqrt{A D^{2}-C D^{2}}=2 \sqrt{49-4}=2 \sqrt{45}=6 \sqrt{5}, \\
S_{\triangle A B C}=S_{\triangle A B D}=\frac{1}{2} A B \cdot C K=\frac{1}{2} \cdot 12 \cdot \sqrt{13}=6 \sqrt{13} .
\end{gathered}
$$

Therefore,

$$
r=\frac{3 V}{S}=\frac{72}{12 \sqrt{5}+12 \sqrt{13}}=\frac{6}{\sqrt{5}+\sqrt{13}}
$$

Let $\alpha$ be the angle between the faces $A B D$ and $A B C$. Then

$$
\sin \alpha=\sin \angle C K D=\frac{12}{13}, \cos \alpha=\frac{5}{13}, \sin \frac{12}{2}=\sqrt{\frac{1-\cos \cos _{2}^{2}}{2}}=\frac{2}{\sqrt{13}}
$$

and since the sphere is inscribed in the dihedral angle between these faces, its center $O$ lies in the bisector plane of the dihedral angle, so if $d$ is the distance from point $O$ to edge $A B$, then

$$
d=\frac{\varphi}{\sin \frac{\pi}{2}}=\frac{\frac{6}{\sqrt{5}+\sqrt{13}}}{\frac{2}{\sqrt{13}}}=\frac{3 \sqrt{13}}{\sqrt{5}+\sqrt{13}}
$$