Question:
If the plane $2 x-y+2 z+3=0$ has the distances
$\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4 x-2 y+4 z+\lambda=0$
and $2 x-y+2 z+\mu=0$, respectively, then the maximum value of $\lambda+\mu$ is equal to :
Correct Option: , 3
Solution:
$4 x-2 y+4 z+6=0$
$\frac{|\lambda-6|}{\sqrt{16+4+16}}=\left|\frac{\lambda-6}{6}\right|=\frac{1}{3}$
$|\lambda-6|=2$
$\lambda=8,4$
$\frac{|\mu-3|}{\sqrt{4+4+1}}=\frac{2}{3}$
$|\mu-3|=2$
$\mu=5,1$
$\therefore$ Maximum value of $(\mu+\lambda)=13$.
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