1. Let the point of intersection of the two lines be $(x,y)$.

   The given lines are $ax+2by+3b=0 \quad ...(1)$ $bx-2ay-3a=0 \quad ...(2)$

2. We need the line parallel to the $x$-axis passing through their intersection point.

   A line parallel to the $x$-axis has equation $y=\text{constant}$ So we only need the $y$-coordinate of the intersection point.

3. Solve for $y$.

   Multiply equation (1) by $a$: $a^2x+2aby+3ab=0 \quad ...(3)$

   Multiply equation (2) by $b$: $b^2x-2aby-3ab=0 \quad ...(4)$

4. Add (3) and (4): $a^2x+b^2x=0$ $x(a^2+b^2)=0$

   Since $(a,b)\neq(0,0)$, we have $a^2+b^2\neq 0$ Hence, $x=0$

5. Substitute $x=0$ into equation (1): $2by+3b=0$ $b(2y+3)=0$

   Also, substituting into equation (2): $-2ay-3a=0$ $a(2y+3)=0$

   Since $a$ and $b$ are not both zero, the common relation is $2y+3=0$ $y=-\frac{3}{2}$

6. Therefore, the required line is $y=-\frac{3}{2}$

   This is a line parallel to the $x$-axis, lying below the $x$-axis at a distance $\left| -\frac{3}{2} \right|=\frac{3}{2}$

7. Checking options:

   • A: below the $x$-axis at a distance of $\frac{3}{2}$ ✔️
   • B: below the $x$-axis at a distance of $\frac{2}{3}$ ✘
   • C: above the $x$-axis at a distance of $\frac{3}{2}$ ✘
   • D: above the $x$-axis at a distance of $\frac{2}{3}$ ✘

Hence, the correct option is A.