(c) The general term in the expansion of ${\left( {\frac{3}{2}{x^2} – \frac{1}{{3x}}} \right)^9}$ is ${T_{r + 1}} = {\,^9}{C_r}{\left( {\frac{3}{2}{x^2}} \right)^{9 – r}}{\left( { – \frac{1}{{3x}}} \right)^r}$$ = {\,^9}{C_r}{\left( {\frac{3}{2}} \right)^{9 – r}}\left( { – \frac{1}{3}} \right){x^{18 – 3r}}$
……$(i)$

Now, the coefficient of the term independent of $x$ in the expansion of $(1 + x + 2{x^3})\,\,{\left( {\frac{3}{2}{x^2} – \frac{1}{{3x}}} \right)^9}$ ……$(ii)$

= Sum of the coefficient of the terms ${x^0},{x^{ – 1}}$and ${x^{ – 3}}$ in ${\left( {\frac{3}{2}{x^2} – \frac{1}{{3x}}} \right)^9}$.

For ${x^0}$ in $(i)$ above, $18 – 3r = 0 \Rightarrow r = 6$. For ${x^{ – 1}}$ in

$(i)$ above, there exists no value of $r$ and hence no such term exists. For ${x^{ – 3}}$in 

$(i)$,$18 – 3r = – 3 \Rightarrow r = 7$

$\therefore $ For term independent of $x$, in $(ii)$ the coefficient

$ = 1 \times {\,^9}{C_6}{( – 1)^6}{\left( {\frac{3}{2}} \right)^{9 – 6}}{\left( {\frac{1}{3}} \right)^6} + 2 \times {\,^9}{C_7}{( – 1)^7}{\left( {\frac{3}{2}} \right)^{9 – 7}}{\left( {\frac{1}{3}} \right)^7}$

$ = \frac{{9.8.7}}{{1.2.3}}.\frac{{{3^3}}}{{{2^3}}}.\frac{1}{{{3^6}}} + 2\frac{{9.8}}{{1.2}}( – 1)\frac{{{3^2}}}{{{2^2}}}.\frac{1}{{{3^7}}}$$ = \frac{7}{{18}} – \frac{2}{{27}} = \frac{{17}}{{54}}$.