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

$\begin{array}{l}\frac{\mathrm{x}+3}{3}=\frac{\mathrm{y}+2}{1}=\frac{1-\mathrm{z}}{2}=\mathrm{\lambda }\\ \Rightarrow \text{ point }(3\mathrm{\lambda }-3,\mathrm{\lambda }-2,-2\mathrm{\lambda }+1)\text{ from the plane}\\ \mathrm{substituting} \mathrm{in} \mathrm{theplane} \mathrm{x}+\mathrm{y}+\mathrm{z}=2. \\ 3\lambda -3+\lambda -2+1-2\lambda =2\\ \mathrm{\lambda }=3\end{array}$

$P(6,1,-5)$

$\mathrm{the} \mathrm{distance} \mathrm{of} \text{the point }\mathrm{P}\text{ from the plane }3\mathrm{x}-4\mathrm{y}+12\mathrm{z}=32$
 

$q=\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\frac{78}{13}=6$

Distance = 6 

q = 6, 2q = 12
$\Rightarrow \text{ equation is }{\mathrm{x}}^2-(6+12)\mathrm{x}+6\times 12=0$