Let P be the point of intersection of the line (x + 3)/(3) = (y +...

Given,

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

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

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

Since point lie on the plane also,

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

$⇒ 2\lambda = 6$

$⇒ \lambda = 3$

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,

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

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

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

$⇒ {x}^{2} - 18x + 72 = 0$