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 $2x-y+2z+\mu =0,$ respectively, then the maximum value of $\lambda +\mu$ is equal to :

• A. $5$
• B. $9$
• C. $13$
• D. $15$

Answer 1

The correct option is C $13$

Given equation of planes are:-
$2x-y+2z+3=0...\left(1\right)$
$4x-2y+4z+\lambda =0...\left(2\right)$
$2x-y+2z+\mu =0...\left(3\right)$
As the plane $\left(1\right)$ and $\left(2\right)$ are parallel and distance between them is equal to $\frac{1}{3}.$
$\Rightarrow \frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}=\frac{|\lambda -6|}{\sqrt{16+4+16}}$
$\Rightarrow \left|\frac{\lambda -6}{6}\right|=\frac{1}{3}$
$\Rightarrow |\lambda -6|=2$
$\Rightarrow \lambda =8,4$

Distance between plane $\left(1\right)$ and $\left(3\right)$ is $\frac{2}{3}.$
$\Rightarrow \frac{|\mu -3|}{\sqrt{4+1+4}}=\frac{2}{3}$
$\Rightarrow |\mu -3|=2$
$\Rightarrow \mu =5,1$

$\therefore (\mu +\lambda {)}_{max}=13$