The equation of the plane which is equidistant from lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{1} = \frac{z-1}{-2}$ is $Ax + By + Cz + D = 0$ then $|A + B + C + D| = $ __________.
Step-by-Step Solution
Key Concept: The plane equidistant from two skew lines has a normal vector equal to the cross product of their direction vectors and passes through a point equidistant from both lines.
A plane equidistant from two skew lines must be parallel to both lines' direction vectors. The normal to the plane is perpendicular to both direction vectors $\vec{d_1} = (1,2,3)$ and $\vec{d_2} = (3,1,-2)$. Computing $\vec{n} = \vec{d_1} \times \vec{d_2} = (-7, 11, -5)$, giving plane equation $-7x + 11y - 5z + D = 0$. The plane passes through the midpoint of the closest points on both lines. Using the condition that distances from both lines are equal and solving for D, we get $D = -6$. Therefore $|A+B+C+D| = |-7+11-5-6| = |-7| = 7$.
Correct Answer: 7