The distance between the planes $2x+3y +4z=4$ and $4x+6y+8z=12$ is :

$\frac{2}{\sqrt{29}} $units

Planes: $2x+3y+4z=4$ and $4x+6y+8z=12$

Divide second equation by $2$: $2x+3y+4z=6$

So, parallel planes: $2x+3y+4z=4$ and $2x+3y+4z=6$

Distance between parallel planes $ax+by+cz=d_1$ and $ax+by+cz=d_2$ is

$D=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$

$=\frac{|4-6|}{\sqrt{2^2+3^2+4^2}}=\frac{2}{\sqrt{29}}$

Required distance = $\frac{2}{\sqrt{29}}$