Given lines $ax+2by+3b=0$

                  $bx-2ay-3a=0$

Point of intersection of the lines is

$\begin{array}{l}x y 1\\ \begin{array}{cccc}2b& 3b& a& 2b\\ -2a& -3a& b& -2a\end{array}\end{array}$

$\Rightarrow \frac{x}{-6ab+6ab}=\frac{y}{3b^2+3a^2}=\frac{1}{-2a^2-2b^2}$

$\begin{array}{l}\Rightarrow \frac{x}{0}=\frac{y}{3\left(a^2+b^2\right)}=\frac{1}{-2\left(a^2+b^2\right)}\\ \Rightarrow \frac{x}{0}=\frac{y}{3}=\frac{1}{-2}\end{array}$

$\therefore \left(x,y\right)=\left(0,\frac{-3}{2}\right)$

Equation of line parallel to x-axis and passing through

$\left(0,\frac{-3}{2}\right)=\left(x_{1 },y_1\right)$

is $y=y_1$

$\Rightarrow y=\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$