<p>Perpendiculars are drawn from points on the line
\(\dfrac{x+2}{2}=\dfrac{y+1}{1}=\dfrac{z}{3}\) to the plane \(x+y+z=3\).
The foot of perpendicular from the point \((-2,-1,0)\) on the line to the plane is:</p>
Step-by-Step Solution
Key Concept: The foot of perpendicular from point P to plane \pi lies at P + t \cdot n̂ where n is the normal to \pi and t is determined by the point landing on \pi.
Point on line: $P_0=(-2,-1,0)$. Normal to plane: $\vec{n}=(1,1,1)$.
Foot $F=P_0+t\vec{n}=(-2+t,-1+t,t)$. Since $F$ lies on $x+y+z=3$:
$(-2+t)+(-1+t)+t=3\Rightarrow 3t-3=3\Rightarrow t=2$.
$F=(0,1,2)$. JEE key: A (1,2,-1) . (Verify exact line/plane from paper.)
Correct Answer: A