Define $f(z)=\frac{1}{z(z-1)(z-2)}$ then we have $\forall r\in(0,1) :\; I_{r}=2\pi iRes[f;0]$ $\forall r\in(1,2) :\; I_{r}=2\pi i(Res[f;0]+Res[f,1])$ $\forall r\in (2,\infty) :\; I_{r}=2\pi i(Res[f;0]+Res[f,1]+Res[f;2])$ Now because $f(z)=\frac{1}{z(z-1)(z-2)}$ is at shape $f(z)=\frac{p(z)}{q(z)}$ when $p(z)=1$ and $q(z)=z(z-1)(z-2)$ that both are analytic in points $z_{0}=0,1,2$ and $p(z_{0})\neq 0$ , $q(z_{0})=0$ and $q'(z_{0})\neq 0$ for these three points. So $f$ has three simple pole there. And so $Res[f;z_{0}]=lim_{z\longrightarrow z_{0}}(z-z_{0})f(z)$ in those points. $Res[f;0]=lim_{z\longrightarrow 0}(z)\frac{1}{z(z-1)(z-2)}=\lim_{z\longrightarrow 0}\frac{1}{(z-1)(z-2)}=\frac{1}{2}$ $Res[f;1]=lim_{z\longrightarrow 1}(z-1)\frac{1}{z(z-1)(z-2)}=\lim_{z\longrightarrow 1}\frac{1}{z(z-2)}=-1$ $Res[f;2]=lim_{z\longrightarrow 2}(z-2)\frac{1}{z(z-1)(z-2)}=\lim_{z\longrightarrow 2}\frac{1}{z(z-1)}=\frac{1}{2}$ And at the end; $\forall r\in(0,1) :\; I_{r}=\pi i$ $\forall r\in(1,2) :\; I_{r}=-\pi i$ $\forall r\in(2,\infty) :\; I_{r}=0$ So the last choice "d" is correct choice.