JEE Main 2019 · 10 April · Shift II

JEE Main 2019 Mathematics question on Three Dimensional Geometry

If the plane 2x−y+2z+3=02x - y + 2z + 3 = 0 has the distances 13\frac{1}{3} and 23\frac{2}{3} units from the planes 4x−2y+4z+λ=04x - 2y + 4z + \lambda = 0 and 2x−y+2z+μ=02x - y + 2z + \mu = 0, respectively, then the maximum value of λ+μ\lambda + \mu is equal to
  1. A.
    13
  2. B.
    15
  3. C.
    5
  4. D.
    9
Show answer

Correct option: A

[JEE Main 2019 · 10 April · Shift II]

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