Given,

$P$ be the point of intersection of the line $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ and the plane $x+y+z=2$

Now let, $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=\lambda$

Say, point $A\left(3\lambda -3, \lambda -2, 1-2\lambda \right)$

Since point lie on the plane also,

So, $3\lambda -3+\lambda -2+1-2\lambda =2$

$\Rightarrow 2\lambda =6$

$\Rightarrow \lambda =3$

Hence, the point will be $P\left(6, 1, -5\right)$

Now finding the distance of the point $P\left(6, 1, -5\right)$ from the plane $3x-4y+12z=32$ we get,

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

Equation with roots $q$ and $2q$ will be,

$x^2-3qx+2q^2=0$

$\Rightarrow x^2-18x+72=0$