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

Line: $$\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2} = t$$

Parametric: $$x = 3t-3, y = t-2, z = 1-2t$$

Substituting in plane $$x + y + z = 2$$:

$$(3t-3) + (t-2) + (1-2t) = 2$$

$$2t - 4 = 2$$

$$t = 3$$

$$P = (6, 1, -5)$$

Distance from $$3x - 4y + 12z = 32$$:

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

So $$q = 6, 2q = 12$$.

Equation with roots 6 and 12: $$x^2 - 18x + 72 = 0$$

The correct answer is Option 2.