jmc

Let $n-1$, $n$, and $n+1$ be three consecutive integers. We have $(n-1)+n+(n+1)=3n=89a$ for some positive integer $a$. Since $(3,89)=1$, $a$ must be a multiple of 3, say $a=3b$ for a positive integer $b$. We must have $600\le 89\cdot 3b\le 900\Rightarrow 600\le 267b\le 900\Rightarrow 2<b<4$. Therefore, $b=3$, which means $3n=267\cdot 3=\boxed{801}$.