Let $12n^2+12n+11=1111k$. Taking mod $4$, $3$, we get $k \equiv 1\pmod{4}, k \equiv 2 \pmod{3}$. This means $k$ must be $5$. So, we want to solve the quadratic $$12n^2+12n-5544=0 \rightarrow n^2+n-462=0$$$$\rightarrow \boxed{n = 21, -22}$$