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Question #41617

Difficulty: 8/10 | 🟢 Free

The Question

Let A=(120011101)A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} and B=(101110021)B = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix}. Let M=adj(AB)M = \text{adj}(AB). Consider the system of linear equations MX=(1−12)M X = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, where X=(xyz)X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}. Let P1P_1 and P2P_2 be the planes represented by the first and second equations of this system, respectively. If LL is the line of intersection of the planes P1P_1 and P2P_2, and LL intersects the plane 2x−3y+z=32x - 3y + z = 3 at the point Q(α,β,γ)Q(\alpha, \beta, \gamma), then the value of 9(α+β+γ)9(\alpha + \beta + \gamma) is:

Options

  • A.

    78

  • B.

    52

  • C.

    64

  • D.

    86

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Line-Plane intersection logic|System of Linear Equations (Cramer/Inverse)|Minors and Cofactors Sign Convention|Matrix Multiplication Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of systems of linear equations and properties of the adjugate matrix.

The primary trap targets the non-commutative property of matrix multiplication.

Students often incorrectly assume adj(AB)=adj(A)adj(B)\text{adj}(AB) = \text{adj}(A)\text{adj}(B) instead of the correct reversal property adj(B)adj(A)\text{adj}(B)\text{adj}(A).

If they use the incorrect order, they effectively solve the system for BABA instead of ABAB, leading to a completely different line of intersection and the trap answer 52.

Additionally, the problem rewards deep conceptual understanding: recognizing that the cross product of the first two rows of adj(K)\text{adj}(K) is proportional to the third column of KK, and that the intersection point can be found via X=1∣K∣KCX = \frac{1}{|K|} K C, bypassing tedious cofactor calculations.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptLine-Plane intersection logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree