The term independent of x in $\left(1+x+2x^2\right){\left(\frac{3x^2}{2}-\frac{1}{3x}\right)}^9$ is

In the expansion ${\left(\frac{3x^2}{2}-\frac{1}{3x}\right)}^9$

$T_{r+1}={}^9{\text{C}}_r{\left(\frac{{\displaystyle 3x^2}}{{\displaystyle 2}}\right)}^{9-r}{\left(\frac{{\displaystyle 1}}{{\displaystyle 3x}}\right)}^r$

$={}^9{\text{C}}_3{\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}\right)}^{9-r}{\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)}^r{x}^{18-3r}$

$\text{if }18-3r=0\Rightarrow r=6$

$\text{if }18-3r=-1\Rightarrow r=\frac{19}{3}$

$\text{if }18-3r=-2\Rightarrow r=\frac{20}{3}$

$\therefore$The term independent of x in ${\left(\frac{3x^2}{2}-\frac{1}{3x}\right)}^9$  is ${}^9{\text{C}}_6{\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}\right)}^3{\left(-\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)}^6$

But there are no terms contain $\frac{1}{x}\&\frac{1}{x^2}$

$\therefore {}^9{\text{C}}_6.\frac{1}{6^3}=\frac{7}{18}$