smc

Add the two equations. $2a^2+2b^2+2c^2-2ab-2ac-2bc=14$. Now, this can be rearranged and factored. $(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=14$ $(a-b{)}^2+(a-c{)}^2+(b-c{)}^2=14$ $a$, $b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. We see that the only is possibility is $14=9+4+1$. $(a-c{)}^2=9\Rightarrow a-c=3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b=1$ or $2$, or whether $b-c=1$ or $2$. We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a,b,c)=(a,a-1,a-3)$ or $(a,a-2,a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$. $a^2-(a-1{)}^2-(a-3{)}^2+a(a-1)=2011$, after some algebra, simplifies to $7a=2021$. $2021$ is not divisible by $7$, so $a$ is not an integer. The other case gives $a^2-(a-2{)}^2-(a-3{)}^2+a(a-2)=2011$, which simplifies to $8a=2024$. Thus, $a=253$ and the answer is $\boxed{253}$.