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

$\sqrt{6}$ unit 7 unit $\frac{5}{\sqrt{6}}$ unit $\frac{4}{\sqrt{6}}$ unit

$\sqrt{6}$ unit

$P_1 : x + 2y - z - 5 =0$ $P_2 : 2x + 4y - 2z + 2 = 0 ⇒ x+ 2y - z +1 = 0$ 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