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

(A) 2 units

(B) 4 units

(C) 8 units

(D) $\cfrac{2}{{\sqrt {29} }}$ units

Option D is correct

The given planes are $2x + 3y + 4z = 4$ ….(1)

and $4x + 6y + 8z = 12$ …..(2)

Dividing (2) by 2, $2x + 3y + 4z = 6$

Distance between parallel planes $= \left| {\cfrac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$

$= \left| {\cfrac{{4 - 6}}{{\sqrt {4 + 9 + 16} }}} \right| = \cfrac{2}{{\sqrt {29} }}$ units.