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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Three Dimensional Geometry

Let the plane containing the line of intersection of the planes P1 : x+(λ+4)y+z=1x + (\lambda + 4)y + z = 1 and P2 : 2x+y+z=22x + y + z = 2 pass through the points (0,1,0)(0, 1, 0) and (1,0,1)(1, 0, 1). Then the distance of the point (2λ,λ,−λ)(2\lambda, \lambda, -\lambda) from the plane P2 is
  1. A.
    262\sqrt{6}
  2. B.
    565\sqrt{6}
  3. C.
    363\sqrt{6}
  4. D.
    464\sqrt{6}
Show answer and solution

Correct option: C

Solution

The family of planes through the line of intersection is P1 + μP2 = 0:

(1+2μ)x + (λ+4+μ)y + (1+μ)z = 1+2μ

Substituting (0,1,0): (λ+4+μ) = 1+2μ → λ+3 = μ

Substituting (1,0,1): (1+2μ) + (1+μ) = 1+2μ → 2+3μ = 1+2μ → μ = -1

So λ+3 = -1 → λ = -4.

The point is (2λ, λ, -λ) = (-8, -4, 4).

Distance from P2: 2x+y+z=2 is |2(-8)+(-4)+4-2|/√(4+1+1) = |-16-4+4-2|/√6 = |-18|/√6 = 18/√6 = 3√6.

[JEE Main 2023 · 24 Jan · Shift 2]

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