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

The given plane is $2x-y+2z+3=0$

or $4x-2y+4z+6=0$

$\Rightarrow$ Piane.(i) is parallel to the plane

$4x-2y+4z+\lambda =0$

$\therefore$ Distance between these two planes is $\frac{|\lambda -6|}{\sqrt{16+4+16}}=\frac{1}{3}$

$\Rightarrow \frac{|\lambda -6|}{\sqrt{36}}=\frac{1}{3}\Rightarrow |\lambda -6|=\frac{6}{3}=2\Rightarrow \lambda =8,4$

Again distance between $2x-y+2z+3=0$ and $2x-y+2z+\mu =0$

$-\mu =0\text{ is }$

$\frac{|\mu -3|}{\sqrt{4+1+4}}=\frac{2}{3}\Rightarrow \frac{|\mu -3|}{\sqrt{9}}=\frac{2}{3}\Rightarrow |\mu -3|=2\Rightarrow \mu =5,1$

$\therefore$ Maximum value of $\mu +\lambda =13$