If $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}\forall x\in R$, $-\left\{\frac{{\displaystyle -3}}{{\displaystyle k}},\frac{{\displaystyle k}}{{\displaystyle 3}}\right\}$, then both the roots of the equation; $k{x}^2-7x+3=0$ are 

Given $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$
$\Rightarrow k(3x-k)+3(kx+3)=12x+5$
Comparing both sides, we get $k=2$
Hence $k{x}^2-7x+3=0$
$\Rightarrow 2x^2-7x+3=0$
Then $x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,x=3$ 
$\therefore$ roots are rational numbers