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. 15
• B. 5
• C. 13
• D. 9

Answer 1

Answer: (C)

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