If the plane 2x - y + 2z + 3 = 0 has distance | vedclass

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(B) The given plane is $2x - y + 2z + 3 = 0$. First,rewrite the plane $4x - 2y + 4z + \lambda = 0$ as $2x - y + 2z + \frac{\lambda}{2} = 0$. The dista

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$13$

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(B) The given plane is $2x - y + 2z + 3 = 0$. First,rewrite the plane $4x - 2y + 4z + \lambda = 0$ as $2x - y + 2z + \frac{\lambda}{2} = 0$. The distance between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. For the first plane,the distance is $\frac{|\frac{\lambda}{2} - 3|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{1}{3}$. $\frac{|\frac{\lambda - 6}{2}|}{3} = \frac{1}{3} \implies |\lambda - 6| = 2$. Thus,$\lambda - 6 = 2$ or $\lambda - 6 = -2$,which gives $\lambda = 8$ or $\lambda = 4$. For the second plane $2x - y + 2z + \mu = 0$,the distance is $\frac{|\mu - 3|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{2}{3}$. $\frac{|\mu - 3|}{3} = \frac{2}{3} \implies |\mu - 3| = 2$. Thus,$\mu - 3 = 2$ or $\mu - 3 = -2$,which gives $\mu = 5$ or $\mu = 1$. The maximum value of $\lambda + \mu$ is $8 + 5 = 13$.

If the plane $2x - y + 2z + 3 = 0$ has distances of $\frac{1}{3}$ 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:

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