jmc

The sum of the coefficients in $$3(3x^7+8x^4-7)+7(x^5-7x^2+5)$$(or any polynomial) can be found by plugging in $x=1$. Then, we have $$3(3+8-7)+7(1-7+5)=3\cdot 4+7\cdot -1=\boxed{5}.$$