Question

If the plane $2x-y+2z+3=0$ has the distances $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x-2y+4z+\lambda =0$ and $4x-2y+4z+\lambda =0$ and $2x-y+2z+\mu =0$ respectively, then the maximum value of $\lambda +\mu$ is equal to:

• A. $15$
• B. $5$
• C. $13$
• D. $9$

Answer 1

The correct option is C $13$

$4x-2y+4z+6=0$
$\frac{|\lambda -6|}{\sqrt{16+4+16}}=\left|\frac{\lambda -6}{6}\right|=\frac{1}{3}$
$|\lambda -6|=2\Rightarrow \lambda =8,4$
$\frac{|\mu -3|}{\sqrt{4+4+1}}=\frac{2}{3}$
$|\mu -1|=2\Rightarrow \mu =5,1$
maximum value of $(\lambda +\mu )=13$