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Question #41627

Difficulty: 8/10 | 🟢 Free

The Question

Let a moving point $P$ trace a surface $S_1$ such that the difference of the squares of its distances from $A(1, 2, 3)$ and $B(3, 0, 1)$ is $-4$ . Let another moving point $Q$ trace a surface $S_2$ such that the difference of the squares of its distances from $C(2, -1, 4)$ and $D(0, 1, 2)$ is $16$ . A plane $\Pi$ passes through the curve of intersection of $S_1$ and $S_2$ , and is perpendicular to the plane $2x + y + 3z = 5$ . If $(\alpha, \beta, \gamma)$ is the image of the point $M(4, -1, 0)$ in the plane $\Pi$ , find the value of $\alpha^2 + \beta^2 + \gamma^2$ .

Options

A.
29
B.
33
C.
38
D.
41

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Image of a point in a plane|Equation of plane through intersection|Locus of a moving Point

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to synthesize locus definitions with 3D plane geometry.

The primary trap lies in the phrasing 'difference of the squares of its distances' and 'curve of intersection.' Students often incorrectly assume that loci involving squared distances must yield higher-order quadric surfaces (like hyperboloids or spheres) and get bogged down in complex algebra.

However, the $x^2, y^2, z^2$ terms perfectly cancel out, reducing the surfaces $S_1$ and $S_2$ to linear plane equations.

Recognizing this linear reduction is crucial before applying the family of planes concept and the standard point-image formula.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Image of a point in a plane
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free