Given $A(2,-3) B(-2,1)$

The third vertex lies on $2 x+3 y=9$
i.e. $C\left(x, \frac{9-2 x}{3}\right)$
$\therefore$ Let $P(h, k)$ be any point on the required locus i.e $P$ is the centroid of the triangle $A B C$
$$
\begin{aligned}
& \Rightarrow\left(\frac{2-2+x}{3}, \frac{-3+1+\frac{9-2 x}{3}}{3}\right)=(h, k) \\
& \therefore h=\frac{x}{3}, k=\frac{3-2 x}{9}
\end{aligned}
$$

Eliminating $x$ from the above equations
$$
\begin{aligned}
& \Rightarrow 9 k=3-2(3 h) \\
& \Rightarrow 9 \mathrm{k}=3-6 h \\
& \Rightarrow 2 h+3 k=1
\end{aligned}
$$

Hence, the locus of $P(h, k)$ is $2 x+3 y=1$