Use Cauchy's Residue Theorem:$\oint_\gamma f(z)dz=2\pi i\sum_{a_i\in A}\operatorname{Res}_{z=a_i}f(z)$
When $\,A=\,$interior of the rectifiable curve $\,\gamma\,$ which meets no poles of $\,f\,$ .
Note that taking $\,r\in (2,3)\,$ or taking $\,r>3\,$ is the same regarding this integral (why?), and since all the function's poles are simple you can easily calculate its residue at pole $\,a_k\,$ by evaluating $\lim_{z\to a_k}(z-a_k)f(z)$ with $f(z):=\frac{1}{z(z-1)(z-2)}$
Added For any $\,r>0\,\,,\mathcal{C}_r\,$ is a circle centered at the origin and radius $\,r\,$, thus for instance:
$\,(2)\,$ For $\,r\in (1,2)\,\,,\,\mathcal{C}_r\,$ is a circle centered at the origin that intersects the $x-$axis at some point between $\,1\,$ and $\,2\,$, thus the inner part of this circle, $\,A\,$ (which is inclosed by the path $\,|z|=r\,$ , the circle's perimeter if you will) only contains the poles $\,0,1\,$of the function $\,f(z)\,$, and thus here $I_r=2\pi i\sum_{a_i\in A}\operatorname{Res}_{z=a_i}f(z)=2\pi i\left(\frac{1}{2}+(-1)\right)=-\pi i$ Why? Because for example, as stated above: $\operatorname{Res}_{z=1}f(z)=\lim_{z\to 1}\left[(z-1)\frac{1}{z(z-1)(z-2)}\right]=\frac{1}{1\cdot (1-2)}=-1$
Similarly, the residue at $\,z=0\,$ equals $\,1/2\,$, as you can readily check, and now you can try the other options...
Ps. The formula above to evaluate the residues works for simple poles ...!