Distance between the two planes:

The equations of the planes are

$2 x+3 y+4 z=4$                                ...(1)

$4 x+6 y+8 z=12$

$\Rightarrow 2 x+3 y+4 z=6$            ...(2)   

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, $a x+b y+c z=d_{1}$ and $a x+b y+c z=d_{2}$, is given by,

$D=\left|\frac{d_{2}-d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$

$\Rightarrow D=\left|\frac{6-4}{\sqrt{(2)^{2}+(3)^{2}+(4)^{2}}}\right|$

$D=\frac{2}{\sqrt{29}}$

Thus, the distance between the lines is $\frac{2}{\sqrt{29}}$ units.

Hence, the correct answer is D.