JEE Main 2019 — Three Dimensional Geometry Question with Solution

The distance between the parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is $\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}.$

Given, the distance between the planes $2x-y+2z+3=0$  and $4x-2y+4z+\lambda =0, \Rightarrow 2x-y+2z+\frac{\lambda }{2}=0$ is $\frac{1}{3}$ units

$\Rightarrow \frac{\left|{\displaystyle \frac{\lambda }{2}}-3\right|}{\sqrt{4+1+4}}=\frac{1}{3}$

$\Rightarrow \frac{{\displaystyle \left|\frac{\lambda }{2}-3\right|}}{3}=\frac{1}{3}$

$\Rightarrow \left|\frac{\lambda }{2}-3\right|=1$

$\Rightarrow \frac{\lambda }{2}-3=1$ or $\frac{\lambda }{2}-3=-1$

$\Rightarrow \lambda =8$ or $\lambda =4.$

Also, distance between the planes $2x-y+2z+3=0$ and $2x-y+2z+\mu =0$ is $\frac{2}{3}$

$\Rightarrow \frac{\left|\mathrm{\mu }-3\right|}{\sqrt{4+1+4}}=\frac{2}{3}$

$\Rightarrow \frac{\left|\mathrm{\mu }-3\right|}{3}=\frac{2}{3}$

$\Rightarrow \left|\mathrm{\mu }-3\right|=2$

$\Rightarrow \mathrm{\mu }-3=2$ or $\mathrm{\mu }-3=-2$

$\Rightarrow \mathrm{\mu }=5$ or $\Rightarrow \mathrm{\mu }=1.$

Hence, the maximum value of $\lambda +\mu =8+5=13.$