jmc

Note that $\frac{a}{25-a}+1=\frac{a}{25-a}+\frac{25-a}{25-a}=\frac{a+25-a}{25-a}=\frac{25}{25-a}$. The same trick can be used with the other two terms, so $\frac{b}{65-b}+1=\frac{65}{65-b}$, and $\frac{c}{60-c}+1=\frac{60}{60-c}$. Thus, we add 1 to each term on the left side of our equation: $$\frac{a}{25-a}+1+\frac{b}{65-b}+1+\frac{c}{60-c}+1=7+1+1+1.$$ Now we can use the substitution we derived earlier, so $$\frac{25}{25-a}+\frac{65}{65-b}+\frac{60}{60-c}=10.$$ Finally, we divide everything by $5$ to find that $$\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c}=\boxed{2}.$$