Since $P(2,3)$ and $Q(-1,2)$ are conjugate point w.r.t. the circle $x^2+y^2+2 g x+3 y-2=0$
Hence both the points lie on the polar of the other point
Let $P\left(x_1, y_1\right)=P(2,3)$
$$
Q\left(x_2, y_2\right)=Q(-1,2)
$$

Hence
$$
\begin{aligned}
& \Rightarrow\left(x_1 x_2\right)+\left(y_1 y_2\right)+2 g\left(\frac{x_1+x_2}{2}\right)+3\left(\frac{y_1+y_2}{2}\right)-2=0 \\
& \Rightarrow g=-\frac{19}{2}
\end{aligned}
$$
Since radius $r=\sqrt{g^2+f^2-c}$
$$
r=\sqrt{\left(-\frac{19}{2}\right)^2+\left(-\frac{3}{2}\right)^2-(-2)}=\frac{3 \sqrt{21}}{\sqrt{2}}
$$