If $a\ne 0$ and the line $2bx+3cy+4d=0$ passes through the points of intersection of the parabolas $y^2=4ax$ and $x^2=4ay,$ then

The points of intersection of the parabolas $y^2=4ax$ and $x^2=4ay$ are $(0,0)$ and $(4a,4a)$.
Substituting $(0,0)$ into the line $2bx+3cy+4d=0$:
$2b(0)+3c(0)+4d=0\Rightarrow 4d=0\Rightarrow d=0$.Substituting $(4a,4a)$ into the line:$2b(4a)+3c(4a)+4d=0$.Since $d=0$:$8ab+12ac=0$.Dividing by $4a$ (given $a\ne 0$):$2b+3c=0$.
Therefore, both $d=0$ and $2b+3c=0$ must hold, which means $d^2+(2b+3c{)}^2=0$.