Distance between the parallel planes $2x-y+3z+4=0$ and $6x-3y+9z-3=0$ is:

The perpendicular distance from the origin to the plane $ax+by+cz+d=0$ is 

given by $\frac{d}{\sqrt{a^2+b^2+c^2}}$

Thus the perpendicular distance from origin to the plane $2x-y+3z+4=0$ is

$P_1=\frac{4}{\sqrt{14}}$

Similarly, $P_2=\frac{-3}{3\sqrt{14}}=\frac{-1}{\sqrt{14}}$

Thus the distance between the two parallel planes 

$=\left|{\mathrm{P}}_2-{\mathrm{P}}_1\right|=\frac{5}{\sqrt{14}}$.