Equation of any plane through the line of intersection of the given two planes
$ax+by+cz+d=0$ and $a'x+b'y+c'z+d'=0$ is given by
$(ax+by+cz+d)+\lambda(a'x+b'y+c'x+d')=0$
$\Rightarrow\:(a+\lambda a')x+(b+\lambda b')y+(c+\lambda c')z+(d+\lambda d')=0$
But given that this plane is parallel to the line $y=0,\:z=0$ (which is $x \:axis$).
$\Rightarrow\:$ The normal $\overrightarrow n=(a+\lambda a',b+\lambda b',c+\lambda c')$ is $\perp$ to the line.( i.e., $x \:axis.$)
$\Rightarrow\:(a+\lambda a',b+\lambda b',c+\lambda c').(1,0,0)=0$
$\Rightarrow\:a+\lambda a'=0\:\:\:or\:\:\:\lambda=-\large\frac{a}{a'}$
Substituting the value of $\lambda$ we get the required eqn. of the plane is
$(ab'-a'b)y+(bc'-b'c)z+(ac'-a'c)=0$