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Question #41293

Difficulty: 8/10 | 🟢 Free

The Question

Let the lines $L_1: \frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$ and $L_2: \frac{2x-6}{a} = \frac{1-y}{-2} = \frac{z+1}{1}$ be coplanar. Let $V$ be the unique point of intersection of the planes $P_1: x + y + z = a$ , $P_2: 2x - y + z = 3$ , and $P_3: x + 2y - z = 4$ . If $A, B,$ and $C$ are the points of intersection of the plane $x + y + z = 12$ with the $x, y,$ and $z$ axes respectively, find the volume of the tetrahedron $VABC$ .

Options

A.
144
B.
216
C.
288
D.
408

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

Determinant as Geometric Area/Volume|Coplanarity Condition|Concurrent Lines Condition|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This question tests the coplanarity condition of lines and the determinant formulation of a tetrahedron's volume.

The primary trap lies in the non-standard form of the line $L_2$ : students often blindly extract the denominators as direction ratios without making the coefficients of $x, y, z$ unity, leading to incorrect direction vectors like $(a, -2, 1)$ or $(a, 2, 1)$ .

This 'Coefficient Trap' yields $a=-5$ or $a=3$ , corresponding to the incorrect options 408 and 216.

Furthermore, the problem rewards recognizing that the sum of the coordinates of $V$ is directly given by the plane $P_1$ , bypassing the tedious Cramer's rule calculation for $V$ and directly evaluating the volume determinant.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Determinant as Geometric Area/Volume
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free