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(A) The equation of the line is $\frac{x+3}{3} = \frac{y+2}{1} = \frac{z-1}{-2} = \lambda$. Any point $P$ on the line is given by $(3\lambda - 3, \lam

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

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(A) The equation of the line is $\frac{x+3}{3} = \frac{y+2}{1} = \frac{z-1}{-2} = \lambda$. Any point $P$ on the line is given by $(3\lambda - 3, \lambda - 2, 1 - 2\lambda)$. Since $P$ lies on the plane $x + y + z = 2$,we substitute the coordinates: $(3\lambda - 3) + (\lambda - 2) + (1 - 2\lambda) = 2$. $2\lambda - 4 = 2 \Rightarrow 2\lambda = 6 \Rightarrow \lambda = 3$. Thus,the coordinates of $P$ are $(3(3) - 3, 3 - 2, 1 - 2(3)) = (6, 1, -5)$. The distance $q$ of point $P(6, 1, -5)$ from the plane $3x - 4y + 12z - 32 = 0$ is given by: $q = \left| \frac{3(6) - 4(1) + 12(-5) - 32}{\sqrt{3^2 + (-4)^2 + 12^2}} \right| = \left| \frac{18 - 4 - 60 - 32}{\sqrt{9 + 16 + 144}} \right| = \left| \frac{-78}{13} \right| = 6$. So,$q = 6$ and $2q = 12$. The quadratic equation with roots $6$ and $12$ is $(x - 6)(x - 12) = 0$,which simplifies to $x^2 - 18x + 72 = 0$.

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:

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