JEE Main 2025 Apr 3 Shift 1, Mathematics Q2: Algebra Of Determinants

JEE Main2025Apr 3, Shift 1Mathematics

Let a line passing through the point $(4, 1, 0)$ intersect the line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ at the point $A (α, β, γ)$ and the line $L_{2} : x - 6 = y = - z + 4$ at the point $B (a, b, c)$ . Then $101 α β γ ab c$ is equal to

A
8
B
16
C
12
D
6

Solution

Parametrise the two lines: $A = (2 p + 1, 3 p + 2, 4 p + 3)$ on $L_{1}$ and $B = (q + 6, q, 4 - q)$ on $L_{2}$ .

$P = (4, 1, 0)$ , $A$ , $B$ are collinear, so the direction ratios $P A$ and $P B$ are proportional:

$P A = (2 p - 3, 3 p + 1, 4 p + 3), P B = (q + 2, q - 1, 4 - q) .$

Equating the three ratios yields a system whose solution (after eliminating $pq$ ) gives $p = - 1, q = 3$ .

Therefore $A = (- 1, - 1, - 1)$ and $B = (9, 3, 1) .$

Computing the determinant by column reduction ( $C_{3} \to C_{3} - C_{1}$ ):

$101 - 1 - 1 - 1 931 = 100 - 1 - 10 93 - 8 = 1 \cdot (- 1) (- 8) - 0 + 0 = 8.$

Alternatively expanding along the top row gives $1 \cdot (- 1 + 3) - 0 + 1 \cdot (- 3 + 9) = 2 + 6 = 8.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q2: Algebra Of Determinants

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