We have,
$k^{\text {th }}$ term is the expansion of $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^6$ is independent of $x$
$$
\text { Now, } \begin{aligned}
T_{p+1} & ={ }^6 C_p\left(\frac{3}{2}\right)^{6-p}(x)^{12-2^p} \times\left(-\frac{1}{3}\right)^p\left(x^{-p}\right) \\
T_{p+1} & ={ }^6 C_p\left(\frac{3}{2}\right)^{6-p}(x)^{12-3 p}\left(-\frac{1}{3}\right)^{\varphi}
\end{aligned}
$$
$T_{p+i}$ is independent of $x$
$$
\begin{aligned}
& \therefore 12-3 r & =0, r=4 \\
& \therefore k =r+1=5
\end{aligned}
$$


$$
\begin{aligned}
& \text { Greatest term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^5 \\
& r=\left[\frac{n+1}{1+\left|\frac{9 x^3}{2}\right|}\right]=\left[\frac{6}{1+\frac{9}{2} \times \frac{8}{27}}\right] \quad\left[\because x=\frac{2}{3}\right] \\
& r=\left[\frac{18}{7}\right]=2 \\
& \therefore \quad T_3={ }^5 C_2\left(\frac{3}{2} x^2\right)^3\left(-\frac{1}{3 x}\right)^2=10 \times \frac{27}{8} \times \frac{x^4}{9} \\
& T_3=10 \times \frac{27}{8} \times\left(\frac{2}{3}\right)^4 \times \frac{1}{9}=\frac{20}{27}
\end{aligned}
$$