jmc

One way to do this is to find each inverse explicitly: $$\begin{array}{rl}(3^{-1}+5^{-1}+7^{-1}{)}^{-1}& \equiv (4+9+8{)}^{-1}(\mathrm{mod}11)\\ & \equiv 21^{-1}(\mathrm{mod}11)\\ & \equiv 10^{-1}(\mathrm{mod}11)\\ & \equiv \boxed{10}(\mathrm{mod}11).\end{array}$$ Another way to do this is through manipulation: $$\begin{array}{rl}& (3^{-1}+5^{-1}+7^{-1}{)}^{-1}\\ \equiv & (3\cdot 5\cdot 7)(3\cdot 5\cdot 7{)}^{-1}(3^{-1}+5^{-1}+7^{-1}{)}^{-1}\\ \equiv & (3\cdot 5\cdot 7)(3\cdot 5+5\cdot 7+7\cdot 3{)}^{-1}\\ \equiv & 6\cdot (15+35+21{)}^{-1}\\ \equiv & 6\cdot 5^{-1}\\ \equiv & 6\cdot 9\\ \equiv & \boxed{10}(\mathrm{mod}11)\end{array}$$