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Question #41632

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(1, \lambda, (\lambda-2)^2)$ , $B(\lambda, 1, (\lambda-2)^2)$ , $C((\lambda-2)^2, \lambda, 1)$ and $D(1, 1, 1)$ be four distinct points in $\mathbb{R}^3$ . Find the sum of all real values of $\lambda$ for which these points are coplanar.

Options

A.
2
B.
5
C.
6
D.
3

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Collinearity vs Coplanarity|Singular vs Non-Singular Matrices|Minors and Cofactors Sign Convention|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This question targets the 'Cross-Multiplication Trap' and the 'Degenerate Case Trap'.

Students often divide determinant rows by variable expressions like $(\lambda-1)$ without checking if it equals zero, losing the root $\lambda=1$ .

Furthermore, students who correctly solve the determinant equation $\Delta = 0$ to find $\lambda \in \{1, 2, 3\}$ frequently fail to verify the geometric constraint of the problem.

The problem explicitly states the four points must be distinct.

For $\lambda=1$ and $\lambda=3$ , the points coincide (e.g., $A=C$ for $\lambda=3$ ), making them degenerate.

Blindly summing the roots leads to incorrect answers like 5 or 6.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Collinearity vs Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free