Here is some solved problems from IIT-JEE Previous Year Papers.

**Q. [IIT-JEE 2006]**A plane passes through (1,-2,1) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4, then the distance of the plane from the point (1,2,2) is

(a) 0
(b) 1 (c) √2 (d) 2√2

__Solution :__(d) Let equation of the plane through (1,-2,1) be a(x-1) + b(y+2) + c(z-1)=0 ----------------(1)

Since plane (1) is perpendicular to 2x – 2y + z =0
& x – y +2z =4.

∴2a – 2b + c =0&a – b+2c=0

Solving we have a=k, b=k, c=0.

Substituting in (1), we have x + y +
1=0-----------(2)

Let p be perpendicular distance of the plane (2)
from the point (1,2,2). Then p =
(1+2+3)/√(1

^{2}+1^{2})=2√2.**Q. [IIT-JEE 2003]**The value of k ∍ (x-4)/1 = (y-2)/1 = (z-k)/2 lies in the plane 2x - 4y + z = 7, is

(a) 7 (b) -7 (c) no real value (d) 4

__Solution :__(a) The given line passes through the point P(4,2,k) & lies in plane 2x - 4y + z = 7 . So, P must lie on the plane.