The equation of a line on the plane $x + y + z = 1$ such that the line $\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-1}{1}$ and the required line form a plane which is perpendicular to the plane $x + y + z = 1$ is:
\frac{3x+1}{2} = \frac{3y+1}{-1} = \frac{3z+1}{-1}
\frac{3x-1}{-2} = \frac{3y-1}{1} = \frac{3z-1}{1}
\frac{3x-1}{2} = \frac{3y-1}{-1} = \frac{3z-1}{-1}
None of these
Step-by-Step Solution
Key Concept: The intersection point of the given line with the plane, combined with the perpendicular from a point on the line, determines the foot of perpendicular.
For a point on the given line $\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-1}{1} = r$, we have $(2r+1, r+1, r+1)$. Substituting into the plane equation $x + y + z = 1$ gives $(2r+1) + (r+1) + (r+1) = 1$, so $4r + 3 = 1$, yielding $r = -\frac{1}{2}$. This gives the intersection point $Q = (0, \frac{1}{2}, \frac{1}{2})$. The line from $(1, 1, 1)$ perpendicular to the plane has direction ratios $(1, 1, 1)$, so any point on it is $(1+\lambda, 1+\lambda, 1+\lambda)$. This line intersects the plane when $3(1+\lambda) = 1$, giving $\lambda = -\frac{2}{3}$ and the foot of perpendicular as point $R = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$.
Correct Answer: Looking at this problem, I need to find a line on the plane $x + y + z = 1$ such that the plane formed by this line and the given line is perpendicular to $x + y + z = 1$.
**Key Analysis:**
1. The intersection point of the given line with the plane $x + y + z = 1$ is