Let P be the point of intersection of the line ${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$ and the plane $x+y+z=2$. If the distance of the point P from the plane $3x - 4y + 12z = 32$ is q, then q and 2q are the roots of the equation :

Given, equation of line is <br/><br/>$$ \begin{aligned} & \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\\\ & \therefore x=3 k-3, y=k-2, z=1-2 k \end{aligned} $$ <br/><br/>Since, given that $P \equiv(3 k-3, k-2,1-2 k)$ be the point of intersection of the given line and the plane $x+y+z=2$ <br/><br/>$$ \begin{aligned} &\text { So, }(3 k-3)+(k-2)+(1-2 k) =2 \\\\ &\Rightarrow 2 k-4=2 \Rightarrow k =3 \end{aligned} $$ <br/><br/>Thus, $P=(6,1,-5)$ <br/><br/>Now, distance of point $P$ from the plane $3 x-4 y+12 z=32$ is <br/><br/>$$ \begin{array}{rlrl} &q =\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\left|\frac{-78}{\sqrt{169}}\right| \\\\ &\Rightarrow q =\left|\frac{-78}{13}\right|=\frac{78}{13}=6 \\\\ &\therefore q =6 \Rightarrow 2 q=12 \end{array} $$ <br/><br/>Thus, $q$ and $2 q$ are roots of the equation $x^2-18 x+72=0$