Let $$a$$, $$b$$, $$c$$ and $$d$$ be non-zero numbers. If the point of intersection of the lines $$4ax + 2ay + c = 0$$ and $$5bx + 2by + d = 0$$ lies in the fourth quadrant and is equidistant from the two axes then:

We have the two straight-line equations

$$4ax + 2ay + c = 0 \qquad\qquad (1)$$

$$5bx + 2by + d = 0 \qquad\qquad (2)$$

Because both $$a$$ and $$b$$ are non-zero, we can divide each equation by its respective non-zero constant so that the coefficients of $$x$$ and $$y$$ become simpler.

Dividing (1) by $$2a$$ gives

$$2x + y + \frac{c}{2a}=0 \; \; \Longrightarrow \; \; y = -2x - \frac{c}{2a} \qquad (3)$$

Dividing (2) by $$b$$ gives

$$5x + 2y + \frac{d}{b}=0 \; \; \Longrightarrow \; \; 2y = -5x - \frac{d}{b} \; \; \Longrightarrow \; \; y = -\frac52 x - \frac{d}{2b} \qquad (4)$$

The point of intersection satisfies both (3) and (4), so we equate the right-hand sides:

$$-2x - \frac{c}{2a} \;=\; -\frac52 x - \frac{d}{2b}$$

Multiplying every term by $$2$$ to clear the denominators, we obtain

$$-4x - \frac{c}{a} = -5x - \frac{d}{b}$$

Rearranging,

$$(-4x + 5x) = -\frac{d}{b} + \frac{c}{a} \; \; \Longrightarrow \; \; x = \frac{c}{a} - \frac{d}{b}$$

To combine the two fractions we take the common denominator $$ab$$:

$$x = \frac{cb - ad}{ab} \qquad (5)$$

Now we substitute this value of $$x$$ into equation (3) to obtain $$y$$.

Using (3):

$$y = -2x - \frac{c}{2a}$$

Substituting from (5):

$$\begin{aligned} y &= -2\left(\frac{cb - ad}{ab}\right) - \frac{c}{2a} \\ &= \frac{-2(cb - ad)}{ab} - \frac{c}{2a} \end{aligned}$$

To combine the two terms, we again choose the common denominator $$2ab$$:

$$\frac{-2(cb - ad)}{ab} = \frac{-4(cb - ad)}{2ab} = \frac{-4cb + 4ad}{2ab}$$

and

$$-\frac{c}{2a} = \frac{-cb}{2ab}$$

Adding these we get

$$y = \frac{-4cb + 4ad - cb}{2ab} = \frac{-5cb + 4ad}{2ab} \qquad (6)$$

The point lies in the fourth quadrant, so its $$x$$-coordinate is positive and its $$y$$-coordinate is negative. Moreover, the point is equidistant from the two coordinate axes. The distance of a point $$(x,\,y)$$ from the $$y$$-axis is $$|x|$$ and from the $$x$$-axis is $$|y|$$. Equidistance therefore gives the condition

$$|x| = |y|$$

Since the point is in the fourth quadrant, $$x > 0$$ and $$y < 0$$, so $$|x| = x$$ and $$|y| = -y$$. Hence

$$x = -y \qquad (7)$$

Substituting (5) and (6) into (7), we get

$$\frac{cb - ad}{ab} = -\left(\frac{-5cb + 4ad}{2ab}\right)$$

Multiplying both sides by $$2ab$$ to eliminate denominators yields

$$2(cb - ad) = 5cb - 4ad$$

Expanding the left side and then gathering like terms,

$$2cb - 2ad = 5cb - 4ad$$

Bringing all terms to one side,

$$2cb - 2ad - 5cb + 4ad = 0$$

$$(-3cb) + 2ad = 0$$

Multiplying through by $$-1$$ for convenience, we arrive at

$$3cb - 2ad = 0$$

Since multiplication is commutative, $$3cb = 3bc$$, so the final relation is

$$3bc - 2ad = 0$$

This matches Option A.

Hence, the correct answer is Option A.