Probably Number Theory!

We want to investigate $a(b(c+1)+1) = abc + ab + a \equiv 0 \pmod{3}$ Therefore, either $a\equiv 0 \pmod{3}$ or $b(c+1) \equiv 2 \pmod{3} \iff (b,c) \equiv (1,1), (2,0) \pmod{3}$

Now, there are $\frac{2010}{3} = 670$ elements in the set $\{1,2,\ldots, 2010\}$ corresponding to each residue, so we see that each residue has an equal probability of being chosen. It follows that $\text{Prob}\Big(a\equiv 0\Big) = \frac{1}{3} \\ \text{Prob}\Big( (b,c) \equiv (1,1),(2,0) \Big) = \frac{2}{9} \\ \text{Prob}\Big(a\equiv 0 \text{ and } (b,c) \equiv (1,1), (2,0)\Big) = \frac{2}{27}$ so that $\text{Prob}\Big(a\equiv 0 \text{ or } (b,c) \equiv (1,1), (2,0)\Big) = \frac{1}{3} + \frac{2}{9} - \frac{2}{27} = \boxed{\frac{13}{27}}$