JEE Main 2019 (Online) 10th April Evening ShiftQuestion 1
If the plane $2x-y+2z+3=0$ has the distances $\frac13$ and $\frac23$ 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 :
3D Geometry
Solution
Rewrite as $2x-y+2z+\frac\lambda2$: $\frac{|3-\lambda/2|}{3}=\frac13$ gives $\lambda=4,8$. $\frac{|3-\mu|}{3}=\frac23$ gives $\mu=1,5$. Max $\lambda+\mu=13$.