JEE Math Practice Question
If the plane 2x – y + 2z + 3 = 0 has the distances $${1 \over 3}$$ and $${2 \over 3}$$ units from the planes 4x – 2y + 4z + $$\lambda $$ = 0 and 2x – y + 2z + $$\mu $$ = 0, respectively, then the maximum value of $$\lambda $$ + $$\mu $$ is equal to :
- A.13
- B.9
- C.5
- D.15
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