The line parallel to $$x$$-axis and passing through the point of intersection of lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0$$, where $$(a, b) \neq (0, 0)$$ is

The required line is parallel to the $$x$$-axis, so we only need the $$y$$-coordinate of the point of intersection of the two given lines.

Lines:
$$ax + 2by + 3b = 0$$ $$-(1)$$
$$bx - 2ay - 3a = 0$$ $$-(2)$$

Rewrite each in the form $$px + qy = r$$:

From $$(1): \; ax + 2by = -3b$$ $$-(3)$$
From $$(2): \; bx - 2ay = 3a$$ $$-(4)$$

Multiply $$(3)$$ by $$b$$ and $$(4)$$ by $$a$$ so that the $$x$$ terms become equal:

$$abx + 2b^{2}y = -3b^{2}$$ $$-(5)$$
$$abx - 2a^{2}y = 3a^{2}$$ $$-(6)$$

Subtract $$(6)$$ from $$(5)$$ to eliminate $$x$$:

$$(abx + 2b^{2}y) - (abx - 2a^{2}y) = -3b^{2} - 3a^{2}$$
$$2b^{2}y + 2a^{2}y = -3(b^{2} + a^{2})$$

Factor out $$2(a^{2}+b^{2})$$ on the left (note that $$(a,b)\neq(0,0)\Rightarrow a^{2}+b^{2}\neq0$$):

$$2(a^{2}+b^{2})y = -3(a^{2}+b^{2})$$

Divide both sides by $$2(a^{2}+b^{2})$$:

$$y = -\frac{3}{2}$$

Thus the intersection point has $$y$$-coordinate $$-\dfrac{3}{2}$$, independent of $$a$$ and $$b$$. A line parallel to the $$x$$-axis through this point is

$$y = -\frac{3}{2}$$

The constant $$y$$ is negative, so the line lies below the $$x$$-axis. Its perpendicular distance from the $$x$$-axis is $$\left|\,-\dfrac{3}{2}\,\right| = \dfrac{3}{2}$$.

Therefore, the correct description is: below $$x$$-axis at a distance $$3/2$$ from it.
Option C which is: below $$x$$-axis at a distance $$3/2$$ from it.