Question

Let the projection of the line $L:\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z-2}{3}$ in the plane $3x+2y+z=5$ is $L^{\prime }$. If the distance of a plane from the origin which contains the lines $L$ and $L^{\prime }$ is $p$, then the value of $6p^2$ is

Answer 1

Required plane is $PL{L}^{\prime }$ which also contains a line with d.r's $(2,-1,3).$
Since, required plane is perpendicular to $3x+2y+z=5$
$\therefore$ Required plane is
$\left|\begin{array}{ccc}x-1& y+2& z-2\\ 2& -1& 3\\ 3& 2& 1\end{array}\right|=0$
$\Rightarrow x-y-z-1=0$
Distance of the plane from origin is $\frac{1}{\sqrt{3}}.$

Alternate Solution:
$L:\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z-2}{3}$
$\Rightarrow \overrightarrow{a}=(1,-2,2),\overrightarrow{b}=(2,-1,3)$
$\Rightarrow$ The required plane will pass through $(1,-2,2).$
Let the equation of the plane be $a(x-1)+b(y+2)+c(z-2)=0$
Since, the line is contained in the plane
$\therefore 2a-b+3c=0\cdots \left(1\right)$

Since, required plane contains its image also.
Hence, required plane is perpendicular to the plane $3x+2y+z=5$
$\therefore 3a+2b+c=0\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$a=-b=-c$
So, the equation of the plane is $x-y-z=1$
Distance of the plane from origin is $\frac{1}{\sqrt{3}}.$