The distance of the point (3, –2, 1) from the plane 2x – y + 2z + 3 = 0 is :

\(\frac{13}{3}\)

Distance of point (x, y, z) from plane Ax + By + Cz = D is

$|\frac{Ax_1 + By_1 + Cz_1 = D}{\sqrt{A^2+B^2}}|$  ....(i)

given points is (3, -2, 1)

x₁ = 3, y₁ = -2, z₁ = 1 & equation of plane is 2x - y + 2z + 3 = 0

2x - y + 2z = -3

-(2x - y + 2z) = 3 ⇒ -2x+ y - 2z = 3

comparing with Ax + By + Cz = D

A = -2, B = 1, C = -2, D = 3

Distance of point form the plane = $|\frac{(-2×3)+(1×-2)+(-2×1)-3}{\sqrt{(-2)^2+(1)^2+(2)^2}}|$

$=|\frac{(-6)+(-2)+(-2)-3}{\sqrt{4+1+4}}|⇒|\frac{-13}{\sqrt{9}}|$

$=|\frac{-13}{3}|=\frac{13}{3}$

Option 2 is correct.