$2x+3y+4z=6$ the distance between parallel planes
$ax+by+cz=d_1$ and $ax+by+cz=d_2$
$\therefore \left|\frac{{\mathrm{d}}_1-{\mathrm{d}}_2}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right|$
The distance between the planes (i) and (iii) is
$$\left|\frac{6-4}{\sqrt{2^2+3^2+4^2}}\right|=\left|\frac{2}{\sqrt{4+9+16}}\right|=\frac{2}{\sqrt{29}}$$
Option (d) is correct.
$2x+3y+4z=6$ the distance between parallel planes
$ax+by+cz=d_1$ and $ax+by+cz=d_2$
$\therefore \left|\frac{{\mathrm{d}}_1-{\mathrm{d}}_2}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right|$
The distance between the planes (i) and (iii) is
$$\left|\frac{6-4}{\sqrt{2^2+3^2+4^2}}\right|=\left|\frac{2}{\sqrt{4+9+16}}\right|=\frac{2}{\sqrt{29}}$$
Option (d) is correct.