Let $f\left(x\right)=x^3+3x^2+6x+2009$ and $g\left(x\right)=\frac{1}{x-f\left(1\right)}+\frac{2}{x-f\left(2\right)}+\frac{3}{x-f\left(3\right)}.$ The number of real solutions of $g\left(x\right)=0$ is 

$f^{\mathrm{\prime }}\left(x\right)=3(x+1{)}^2+3>0\mathrm{\forall }x.$

Thus, $f\left(x\right)$ increases on $R.$ Let $a=f\left(1\right),b=f\left(2\right)$ and $c=f\left(3\right),$ then $a<b<c$ and 

$g\left(x\right)=(x-b)(x-c)+2(x-a)(x-c)+3(x-a)(x-b)$

As $g\left(a\right)>0,g\left(b\right)<0\text{ and }g\left(c\right)>0,g\left(x\right)=0$ has exactly two real solutions.