jmc

We complete the square. First, we factor 3 out of the terms $3x^2-6x$ to get $3(x^2-2x)$. We can square $x-1$ to get $x^2-2x+1$, so $3(x^2-2x)=3[(x-1{)}^2-1]=3(x-1{)}^2-3$, and $$3(x^2-2x)-2=3(x-1{)}^2-3-2=3(x-1{)}^2-5.$$We see that $a=3$, $h=1$, and $k=-5$, so $a+h+k=3+1+(-5)=\boxed{-1}$.