12. Four spheres with radii $6,6,6,7$ are pairwise externally tangent, and they are all internally tangent to a larger sphere. What is the radius of the larger sphere? $\qquad$
Ronggui 14.
Analysis Let the centers of the four spheres be $A, B, C, D$, then $A B=B C=C A=12, D A=D B=D C=13$, which means $A B C D$ forms a regular tetrahedron. Let the center of the sphere to be found be $X$, with radius $r$. By symmetry, $D X \perp$ plane $A B C$, that is, the projection of $X$ on plane $A B C$ is the center $O$ of the equilateral triangle $A B C$. It is easy to know that
$$
O A=4 \sqrt{3}, \quad O D=\sqrt{D A^{2}-O A^{2}}=11
$$

Let $O X=x$, then $A X=\sqrt{O X^{2}+O A^{2}}=\sqrt{x^{2}+48}$.
Since sphere $A$ is internally tangent to sphere $X$, we have $A X=r-6$, that is $\sqrt{x^{2}+48}=r-6$.
Also, $D X=O D-O X=11-x$, and since sphere $D$ is internally tangent to sphere $X$, we have $D X=r-7$, thus $11-x=r-7$.
From the above two equations, we can solve for $x=4, r=14$, that is, the radius of the larger sphere is 14.