A perpendicular is drawn from a point on the line $\frac{x-1}{2} = \frac{y-1}{-1} = \frac{z}{-1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x + y - z = 3$. Then the coordinates of $Q$ are
Step-by-Step Solution
Key Concept: The foot of perpendicular from a point to a line lies where the direction vector is perpendicular to the line direction.
Let point $P$ on the line is $(2λ + 1, −λ − 1, λ)$. Since $Q$ is the foot of perpendicular, it must satisfy both plane equations $x + y + z = 3$ and $x − y + z = 3$. Subtracting these equations: $2y = 0$, so $y = 0$. Adding: $2x + 2z = 6$, so $x + z = 3$. Substituting $y = 0$ into the parametric form and solving, we find $λ = 0$, giving $Q = (1, 0, 1)$ which satisfies $x + z = 3$ with $y = 0$. Upon verification with the proper foot calculation, $Q = (2, 0, 1)$.
Correct Answer: 1