Question

Let $d$ is the distance of a plane $P$ from the point $\left(2 , 1 , 0\right)$ . Plane $P$ passes through $\left(1 , 2 , 1\right)$ and contains a line $x+2y+z=0=2x-y-z-1$ then $27d^{2}$ is

Answer 1

Let equation of the plane is $\left(x \, + \, 2 y \, + \, z\right)+\lambda \, \left(2 x \, - \, y \, - \, z \, - \, 1\right)=0$
since it passes through $\left(1 , \, 2 , \, 1\right)$
$\Rightarrow \, 6 \, + \, \left(\right.-2\left.\right)\lambda \, = \, 0\Rightarrow \, \lambda \, = \, 3 \, $
$\therefore $ equation of plane $P$ is: $7x \, - \, y \, - \, 2z \, - \, 3 \, = \, 0$
$d=\left|\frac{14 - 1 - 3}{\sqrt{49 + 1 + 4}}\right|=\left|\frac{10}{\sqrt{54}}\right|$
$d^{2}=\frac{50}{27}$ ​​​​