We proceed with stars and bars. Let $a,b,c$ be the number of $a,b,c$ there are in the $2010$ letter word. The total number of ways is $a+b+c=2010$ thus there are $\binom{2012}{2}$ ways to choose how many of each letter there are. Now we count how many there are if there are an even number of each letter. Let $a=2x, b=2y, c= 2z$ $$2x+2y+2z=2010$$$$x+y+z=1005$$Thus the total number of ways is $\binom{1007}{2}$ Thus our answer is $\binom{2012}{2}-\binom{1007}{2}=\boxed{1,516,545}$