6. Given four spheres with radii $6,6,6,7$ that are pairwise externally tangent, and all are internally tangent to a larger sphere. Then the radius of the larger sphere is $\qquad$ .
6. 14 .

Let the centers of the four spheres be $A$, $B$, $C$, and $D$. Then
$$
\begin{array}{l}
A B=B C=C A=12, \\
D A=D B=D C=13,
\end{array}
$$

which means that the tetrahedron $A B C D$ forms a regular triangular pyramid.
Let the center of the larger sphere be $X$, with radius $r$.
By symmetry, we know that $D X \perp$ plane $A B C$, meaning that the projection of point $X$ onto plane $A B C$ is the center $O$ of the equilateral $\triangle A B C$.
It is easy to see that $O A=4 \sqrt{3}, O D=\sqrt{D A^{2}-O A^{2}}=11$.
Let $O X=x$. Then
$$
A X=\sqrt{O X^{2}+O A^{2}}=\sqrt{x^{2}+48} \text {. }
$$

Since sphere $A$ is internally tangent to sphere $X$, we have
$$
A X=r-6 \Rightarrow \sqrt{x^{2}+48}=r-6 \text {. }
$$

Also, $D X=O D-O X=11-x$, and since sphere $D$ is internally tangent to sphere $X$, we know $D X=r-7$.
Thus, $11-x=r-7$.
Solving the above two equations, we get $x=4, r=14$.