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If the plane 2 x-y+2 z+3=0 has the distances

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 :

  1. 15

  2. 5

  3. 13

  4. 9


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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