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JEE Advanced Mathematics Three Dimensional Geometry 2026 JEE Advanced 2026 (Paper 1)

JEE Advanced Mathematics Question (2026) — Solution

Question

Let P be the plane such that it contains the straight line x-1 2 = y-3 3 = z+2 1 and is perpendicular to the plane x + 2y + 3z = 4. Let P_1 be the plane which passes through the point (4, 2, 2) and is parallel to P. Then which of the following statements is (are) TRUE?

Options

  1. A. The equation of the plane P is 7x - 5y + z = -10
  2. B. The distance between the planes P and P_1 is 30
  3. C. The distance of the plane P from the origin is 2 3
  4. D. The acute angle between the plane P and the plane 2x + 2y + z = 3 is ^ -1 ( 1 3 3 )

Answer

D. The acute angle between the plane P and the plane 2x + 2y + z = 3 is ^ -1 ( 1 3 3 )

Step-by-step solution

Let the normal vector to the plane P be n . Since the plane P contains the line x-1 2 = y-3 3 = z+2 1 , its normal vector n is perpendicular to the line's direction vector b = 2 i + 3 j + k . Since P is perpendicular to the plane x + 2y + 3z = 4, n is perpendicular to its normal vector n _1 = i + 2 j + 3 k . n = (2 i + 3 j + k ) ( i + 2 j + 3 k ) = i (9-2) - j (6-1) + k (4-3) = 7 i - 5 j + k Plane P passes through the point (1, 3, -2) which lies on the given line. Equation of plane P is 7(x-1) - 5(y-3) + 1(z+2) = 0 7x - 5y + z = -10 Statement (1) is TRUE. The plane P_1 is parallel to P and passes through (4, 2, 2). Equation of plane P_1 is 7(x-4) - 5(y-2) + 1(z-2) = 0 7x - 5y + z = 20 Distance between parallel planes P and P_1 is d = |20 - (-10)| 7^2 + (-5)^2 + 1^2 = 30 75 = 2 3 . Statement (2) is FALSE. Distance of plane P from the origin is d_0 = |-10| 75 = 2 3 . Statement (3) is FALSE. Angle between the planes 7x - 5y + z + 10 = 0 and 2x + 2y + z - 3 = 0 is given by: = |7(2) + (-5)(2) + 1(1)| 7^2+(-5)^2+1^2 2^2+2^2+1^2 = 5 75 9 = 5 15 3 = 1 3 3 = ^ -1 ( 1 3 3 ). Statement (4) is TRUE.

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