IIT JEE Advanced 2006 Paper 1

The equation of plane which is passing through the point (1, −2, 1) is given by
a(x – 1) + b(y + 2) + c(z – 1) = 0
is perpendicular to the given plane.
2a – 2b + c = 0
a – b + 2c = 0 ⇒ a – b = −2c
−4c + c = 0 ⇒ c = 0
Therefore, a = b.
Hence, the equation of the required plane is x + y + 1 = 0
Therefore, the distance of point (1, 2, 2) is

|1+2+1|

√2

= 2√2