Step-by-Step Solution
Key Concept: To find the line-plane intersection, parametrize the line and solve the plane equation simultaneously.
Let $\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{4} = \eta$. Then $x = 3+2\eta$, $y = 2+3\eta$, $z = 1+4\eta$. For the line to lie on the plane, we need $\frac{x-2}{3} = \frac{y-3}{2} = \frac{z-2}{3}$, which gives $\eta = 1$ and intersection point $(5, 5, 5)$.
Correct Answer: I need to find which plane the intersection point (5, 5, 5) lies on.
Let me check each option by substituting the point (5, 5, 5):
**Option 1:** $3x + y + z = 20$
$3(5) + 5 + 5 = 15 + 5