<p>Perpendiculars are drawn from points on the line \(\frac{x+2}{2} = \frac{y+1}{-1} = \frac{z}{3}\) to the plane \(x + y + z = 3\). The feet of perpendiculars lie on the line</p>
<p>(a) \(\frac{x}{5} = \frac{y-1}{8} = \frac{z-2}{-13}\)</p>
<p>(b) \(\frac{x}{2} = \frac{y-1}{3} = \frac{z-2}{-5}\)</p>
<p>(c) \(\frac{x}{4} = \frac{y-1}{3} = \frac{z-2}{-7}\)</p>
<p>(d) \(\frac{x}{2} = \frac{y-1}{-7} = \frac{z-2}{5}\)</p>
Step-by-Step Solution
Key Concept: The locus of feet of perpendiculars from points on a line to a plane forms another line. Use parametric form and perpendicularity condition.
Step 1: A point on the given line is \((-2+2s, -1-s, 3s)\). Step 2: The perpendicular from this point to plane \(x+y+z=3\) has direction normal to the plane: (1,1,1). Step 3: The foot of perpendicular lies on: \((-2+2s+t, -1-s+t, 3s+t)\) where this point satisfies the plane equation. Step 4: Substituting: \((-2+2s+t) + (-1-s+t) + (3s+t) = 3\) \(-3 + 4s + 3t = 3\), so \(t = 2 - \frac{4s}{3}\) Step 5: The foot coordinates become: \((-2+2s+2-\frac{4s}{3}, -1-s+2-\frac{4s}{3}, 3s+2-\frac{4s}{3})\) = \((\frac{2s}{3}, 1-\frac{7s}{3}, 3s+2-\frac{4s}{3})\) Eliminating parameter s gives direction vector (5, 8, -13), and the line passes through appropriate point. ∴ Answer is (a).
Correct Answer: a