jmc

Let $P=(0,0)$ be at the origin. Using the formula $A=rs$ on $△PQR$, where $r_1$ is the inradius (similarly define $r_2,r_3$ to be the radii of $C_2,C_3$), $s=\frac{PQ+QR+RP}{2}=180$ is the semiperimeter, and $A=\frac{1}{2}bh=5400$ is the area, we find $r_1=\frac{A}{s}=30$. Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles.) Thus $ST,UV$ lie respectively on the lines $y=60,x=60$, and so $RS=60,UQ=30$. Note that $△PQR\sim △STR\sim △UQV$. Since the ratio of corresponding lengths of similar figures are the same, we have $$\frac{r_1}{PR}=\frac{r_2}{RS}⟹r_2=15\text{and}\frac{r_1}{PQ}=\frac{r_3}{UQ}⟹r_3=10.$$ Let the centers of $\odot C_2,C_3$ be $O_2=(0+r_2,60+r_2)=(15,75),O_3=(60+r_3,0+r_3)=(70,10)$, respectively; then by the distance formula we have $O_2{O}_3=\sqrt{55^2+65^2}=\sqrt{10\cdot 725}$. Therefore, the answer is $n=\boxed{725}$.