jmc

We have the factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$Squaring the equation $a+b+c=11,$ we get $$a^2+b^2+c^2+2ab+2ac+2bc=121.$$Then $a^2+b^2+c^2-ab-ac-bc=121-3(ab+ac+bc)=121-75=46,$ so $$a^3+b^3+c^3-3abc=11\cdot 46=\boxed{506}.$$