7. Given that the three sides of $\triangle A B C$ are $13,14,15$. There are 4 circles $O, O_{1}, O_{2}, O_{3}$ with the same radius $r$ placed inside $\triangle A B C$, and circle $O_{1}$ is tangent to sides $A B$ and $A C$, circle $O_{2}$ is tangent to sides $B A$ and $B C$, circle $O_{3}$ is tangent to sides $C B$ and $C A$, and circle $O$ is tangent to circles $O_{1}, O_{2}, O_{3}$. Then $r=$ $\qquad$
7. $\frac{260}{129}$ Detailed Explanation: Let's assume $a=13, b=14, c=15$. We know that $\triangle A B C$ is similar to $\triangle O_{1} O_{2} O_{3}$, and $O$ is the circumcenter of $\triangle O_{1} O_{2} O_{3}$, with the circumradius being $2 r$. Then, $\cos \angle O_{1} O_{3} O_{2}=\cos C=\frac{5}{13}, \sin \angle O_{1} O_{3} O_{2}=\frac{12}{13}$. By the Law of Sines, $O_{1} O_{2}=$ $4 r \sin \angle O_{1} O_{3} O_{2}=\frac{48 r}{13}$. Similarly, we can get $\cos A=\frac{3}{5}, \sin A=\frac{4}{5}, \cos B=\frac{33}{65}, \sin B=\frac{56}{65}$. Also, $O_{1} O_{2}=15-$ $r \cot \frac{A}{2}-r \cot \frac{B}{2}=15-r \frac{1+\cos A}{\sin A}-r \frac{1+\cos B}{\sin B}=15-\frac{15}{4} r$, so $\frac{48 r}{13}=15-\frac{15 r}{4}, r=\frac{260}{129}$.