6. $\odot O_{1} 、 \odot O_{2}$ are externally tangent to each other, their radii are 7 and 14 respectively, the circle $\odot \mathrm{O}_{3}$ that contains these two circles and is tangent to them has its center on their line of centers, the radius of the fourth circle that is tangent to all three circles is $\qquad$.
6. 6 .

It is known that $\odot O_{4}$ is externally tangent to $\odot O_{1}$ and $\odot O_{2}$, and internally tangent to $\odot O_{3}$. Let the radius of $\odot O_{4}$ be $x$, and connect $O_{1} O_{4}$, $O_{2} O_{4}$, and $O_{3} O_{4}$. Then, $O_{1} O_{3}=14$, $O_{3} O_{2}=7$, $O_{1} O_{4}=7+x$, $O_{3} O_{4}=21-x$, and $O_{2} O_{4}=14+x$. In $\triangle O_{1} O_{3} O_{4}$ and $\triangle O_{1} O_{2} O_{4}$, use the cosine rule: $\cos \angle O_{2} O_{1} O_{4}=\frac{(7+x)^{2}+14^{2}-(21-x)^{2}}{2 \cdot 14 \cdot(7+x)}$, $\cos \angle O_{2} O_{1} O_{4}=\frac{(7+x)^{2}+21^{2}-(14+x)^{2}}{2 \cdot 21 \cdot(7+x)}$. Therefore, $\frac{(7+x)^{2}+14^{2}-(21-x)^{2}}{2 \cdot 14 \cdot(7+x)}=$ $\frac{(7+x)^{2}+21^{2}-(14+x)^{2}}{2 \cdot 21 \cdot(7+x)}$, which simplifies to $196 x=1176 \Rightarrow x=6$.