Q49 / 90 · JEE Main 2015 (04 Apr) · Mathematics

The equation of the plane containing the line of intersection of 2x-5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is

  • A
    2 x + 6 y + 1 2 z = - 1 3
  • B
    2 x + 6 y + 1 2 z = 1 3
  • C
    x + 3 y + 6 z = - 7
  • Correct answer
    D
    x + 3 y + 6 z = 7

Solution

Any plane containing the line of intersection of 2x-5y+z=3, x+y+4z=5 will be of the form

2 x - 5 y + z - 3 + λ x + y + 4 z - 5 = 0

2 + λ x - 5 - λ y + 1 + 4 λ z - 3 + 5 λ = 0

It is parallel to plane x + 3 y + 6 z = 1

⇒   2+λ1=λ-53=1+4λ6≠3+5λ1

2 + λ 1 = λ - 5 3    ⇒    6 + 3 λ = λ - 5

2 λ = - 1 1

λ = - 1 1 2

∴ Required plane is

2x-5y+z-3+-112x+y+4z-5=0

2 2 x - 5 y + z - 3 - 1 1 x + y + 4 z - 5 = 0

4 x - 1 0 y + 2 z - 6 - 1 1 x - 1 1 y - 4 4 z + 5 5 = 0

- 7 x - 2 1 y - 4 2 z + 4 9 = 0

⇒     x + 3 y + 6 z - 7 = 0

Hence, the correct answer is option: 'D'.

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