4.11. The centers of three spheres, with radii of 3, 4, and 6, are located at the vertices of an equilateral triangle with a side length of 11. How many planes exist that are tangent to all three spheres simultaneously?

## § 3. Two intersecting circles lie on the same sphere
4.11. Consider a plane tangent to all three given spheres, and draw a plane through the center of the sphere of radius 3 parallel to it. The resulting plane is tangent to the spheres of radii $4 \pm 3$ and $6 \pm 3$, concentric with the spheres of radii 4 and 6. When the signs of the number 3 are the same, the tangency is external, and when they are different, the tangency is internal. It is also clear that for each plane tangent to all spheres, the plane symmetric to it with respect to the plane passing through the centers of the spheres is also tangent to all spheres.

To determine whether there exists a plane passing through a given point and tangent to two given spheres, we can use the result of problem 12.11. In all cases except for the internal tangency with spheres of radii 1 and 9, the tangent planes exist (see Fig. 36). We will prove that there is no plane tangent internally to the spheres of radii 1 and 9 with centers $B$ and $C$ and passing through the point $A$. Let $\alpha$ be the angle between the line $A B$ and the tangent from $A$ to the sphere with center $B$, and $\beta$ be the angle between the line $A C$ and the tangent from $A$ to the sphere with center $C$. It is sufficient to check that $\alpha + \beta > 60^{\circ}$, i.e., $\cos (\alpha + \beta) < 1 / 2$. Since $\sin \alpha = 1 / 11$ and $\sin \beta = 9 / 11$, we have $\cos \alpha = \sqrt{120} / 11$, $\cos \beta = \sqrt{40} / 11$. Then $\cos (\alpha + \beta) = (40 \sqrt{3} - 9) / 121$. Thus, the inequality $\cos (\alpha + \beta) < 1 / 2$ is equivalent to the inequality $80 \sqrt{3} < 139$; the latter inequality is verified by squaring.

In the end, we find that there are 3 pairs of tangent planes.