It looks like there is no restriction on a, b, and c being positive or negative. In that case, the maximum has no specific value. If we let a=n, b=n-1, and c=3 000 001 -2n, then we can let n be as big as we want (a+b+c is still only 3 million). The maximum is ab+1 as n gets bigger (c will be negative), but there is no limit for $n^2-n$ as n approaches infinity. Is the restriction of equality supposed to be something else? Perhaps a, b, c>0?

Or do we want to find the GCD of the three in the parentheses?

I edited problem, thanks.

We can assume that $a>b>c$, let $b-c=x,a-b=y$. We have $3c+2x+y\leq 3000000$. Let $d=\gcd (ab+1,bc+1,ca+1)$, we get $\gcd (d,a)=\gcd (d,b)=\gcd (d,c)=1$. So $d\mid a-c,a-b\Rightarrow d\mid \gcd (x,y)$. Also, $d\mid c^2+1$. We get $d\leq c^2+1,x,y\Rightarrow 3000000\geq 3c+3d\geq 3\sqrt{d-1} +3d$. This gives us $d\leq 999001$. If $d=c^2+1$, we have $d\leq 999^2+1=998002$. If $d= \frac{c^2+1}{2}$, we get $3000000\geq 3\sqrt{2d-1}+3d$. This gives us $d\leq 998587$, so $d\leq \frac{1413^2+1}{2}=998285$. If $d\leq \frac{c^2+1}{5}$, we get $3000000\geq 3\sqrt{5d-1}+3d$. This gives us $d\leq 997766$. So, the maximum value of $d$ is $998285$. Equality holds at $c=1413,x=y=998285$.