Given: Lines   $4ax+2ay+c=0$   and   $5bx+2by+d=0$.

Condition: Intersection lies in the 4th quadrant and is equidistant from the axes $\Rightarrow$ point is of the form $(t,-t)$ with $t>0$.

Substitute $y=-x$ in the two lines:
$4ax+2a(-x)+c=0 \;\Rightarrow\; 2x+\dfrac{c}{2a}=0 \;\Rightarrow\; x=-\dfrac{c}{2a}.$
$5bx+2b(-x)+d=0 \;\Rightarrow\; 3x+\dfrac{d}{b}=0 \;\Rightarrow\; x=-\dfrac{d}{3b}.$

Equate $x$ from both:   $-\dfrac{c}{2a}=-\dfrac{d}{3b} \;\Rightarrow\; \boxed{3bc=2ad}.$

(QIV check: $x>0,\ y<0 \;\Rightarrow\; \dfrac{c}{a}<0$ and $\dfrac{d}{b}<0$ which is consistent.)