Distance between two planes x + 2y – z = 5 and 2x + 4y – 2z + 2 = 0 is

$P_1 : x + 2y - z - 5 =0$

$P_2 : 2x + 4y - 2z + 2 = 0 ⇒ x+ 2y - z +1 = 0$

Both are parallel planes

So distance = $\left|\frac{C_2-C_1}{\sqrt{1^2+2^2+(-1)^2}}\right|=\left|\frac{1+5}{\sqrt{1+4+1}}\right|=\left|\frac{6}{\sqrt{6}}\right|$

$d = \sqrt{6}$ unit

Option: 1