Example 6 Three equal cylindrical surfaces are pairwise tangent, and their axes are perpendicular to each other. If the radius of each cylindrical surface is $r$, find the radius of the smallest sphere that is tangent to these three cylindrical surfaces.

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Solution: It is quite difficult to draw all three cylinders, but their axes are perpendicular and skew, and their distance is $2r$. This geometric structure can be embodied in a cube.

In a cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ with edge length $2a$, $AA_{1}$ lies on the axis of one cylinder, edge $DC$ lies on the axis of another cylinder, and edge $B_{1}C_{1}$ lies on the axis of the third cylinder.

We show that the center of the sphere coincides with the center $O$ of the cube.

In fact, we consider the three cylinders passing through point $O$ and concentric with the given cylinders. Then $O$ is the only point located on or inside all three cylinders, which means that any point other than $O$, i.e., the midpoints of the three lines $AA_{1}$, $DC$, and $B_{1}C_{1}$, is at a greater distance from these three lines than $O$ is. Therefore, it cannot be the center of the sphere tangent to this tri-cylinder. Thus, the radius of the desired sphere is $\sqrt{2}r - r$.