The projection of the line $\frac{x}{5} = \frac{y-1}{2} = \frac{z-1}{1}$ on a plane $P$ is $\frac{x}{1} = \frac{y-1}{1} = \frac{z-1}{-1}$. The plane $P$ passes through $(k, -2, 0)$ then $k = $ __________.
Step-by-Step Solution
Key Concept: Use the point-normal form of a plane after identifying a normal vector from geometric constraints.
The plane passes through the intersection point $(0, 1, 1)$ of two lines. Since the point $(2, 3, 2)$ lies on a line whose projection on the second coordinate axis is $(1, 2, 0)$, the normal vector to the plane is $(1, 1, 2)$. Using the point-normal form, the plane equation becomes $1(x - 0) + 1(y - 1) + 2(z - 1) = 0$, which simplifies to $x + y + 2z - 3 = 0$. Substituting the point $(k, -2, 0)$ into this equation: $k - 2 + 0 - 3 = 0$, so $k = 5$.
Correct Answer: 5