We write this as $(a-b)(a+b)+(a-b)(c)=(a-b)(a+b+c)=2012$. Since $a, b, c$ are positive integers, $a-b<a+b+c$. So, we have three possibilities: $a-b=1$ and $a+b+c=2012$, $a-b=2$ and $a+b+c=1006$, and $a-b=4$ and $a+b+c=503$. The first solution gives $a=b+1$ and $c=2011-2b$, so $b$ can range from 1 through 1005, which determines $a$ and $c$ completely. Similarly, the second solution gives $a=b+2$ and $c=1004-2b$, so $b$ can range from 1 through 501. Finally, the third solution gives $a=b+4$ and $c=499-2b$, so $b$ can range from 1 through 249. Hence, the total number of solutions is $1005+501+249=1755$.