Add the two equations. $$ 2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 a c-2 b c=14 . $$ Now, this can be rearranged and factored.  $$ \begin{aligned} \&\left(a^{2}-2 a b+b^{2}\right)+\left(a^{2}-2 a c+c^{2}\right)+\left(b^{2}-2 b c+c^{2}\right)=14 \textbackslash\textbackslash{} \&(a-b)^{2}+(a-c)^{2}+(b-c)^{2}=14 \end{aligned} $$.  $a, b$, 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 $7 a=2021$. 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 $8 a=2024$. Thus, $a=253$. 
Let $a, b$, and $c$ be positive integers with $a \geq b \geq c$ such that $a^{2}-b^{2}-c^{2}+a b=2011$ and $a^{2}+3 b^{2}+3 c^{2}-3 a b-2 a c-2 b c=-1997$. What is $a$? 
Options:
- $$249$$ 
- $$250$$ 
- $$251$$ 
- $$252$$ 
- $$253$$ 