If the plane 2x-y+2z+3=0 has the distances 1/3 and 2/3 units from the planes 4x-2y+4z+λ=0 and 2x-y+2z+μ=0 respectively, then the maximum value of λ+μ is equal to:
5
13
15
9
Solution:
The correct option is C 4x-2y+4z+6=0 |λ-6|/√16+4+16 = 1/3 |λ-6|/6 = 1/3 |λ-6| = 2 ⇒ λ = 8, 4 |μ-3|/√4+4+1 = 2/3 |μ-3|/3 = 2/3 |μ-3| = 2 ⇒ μ = 5, 1 maximum value of (λ+μ) = 13