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

The two given lines are
$$a x + 2 b y + 3 b = 0 \qquad -(1)$$
$$b x - 2 a y - 3 a = 0 \qquad -(2)$$
with $$(a,b)\neq(0,0)$$.

Let the point of intersection of these lines be $$P(x_0,\,y_0)$$. To find $$P$$ we solve the simultaneous equations $$-(1)$$ and $$-(2)$$.

Write the system in matrix form:
$$\begin{bmatrix} a & 2b \\[2pt] b & -2a \end{bmatrix} \begin{bmatrix} x_0 \\[2pt] y_0 \end{bmatrix} = \begin{bmatrix} -3b \\[2pt] 3a \end{bmatrix}.$$ The determinant of the coefficient matrix is
$$D = a(-2a) - b(2b) = -2(a^2 + b^2) \neq 0.$$ Hence a unique intersection point exists.

Finding $$x_0$$:
Replace the $$x$$-column by the constants and evaluate the determinant: $$D_x = \begin{vmatrix} -3b & 2b \\[2pt] 3a & -2a \end{vmatrix} = (-3b)(-2a) - (2b)(3a) = 6ab - 6ab = 0.$$ Therefore $$x_0 = \frac{D_x}{D} = \frac{0}{-2(a^2 + b^2)} = 0.$$

Finding $$y_0$$:
Replace the $$y$$-column by the constants: $$D_y = \begin{vmatrix} a & -3b \\[2pt] b & 3a \end{vmatrix} = a(3a) - b(-3b) = 3a^2 + 3b^2 = 3(a^2 + b^2).$$ Hence $$y_0 = \frac{D_y}{D} = \frac{3(a^2 + b^2)}{-2(a^2 + b^2)} = -\frac{3}{2}.$$

Thus the intersection point is $$P\bigl(0,\,-\frac{3}{2}\bigr).$$ A line parallel to the $$x$$-axis through this point has equation $$y = -\frac{3}{2}.$$

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

Hence the required description is:
Option A — below the $$x$$-axis at a distance of $$\dfrac{3}{2}$$ from it.