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 \(\mathrm{A}(\alpha, \beta, \gamma)\) and the line \(\mathrm{L}_{2}: x-6=y=-z+4\) at the point \(\mathrm{B}(\mathrm{a}, \mathrm{b}, \mathrm{c})\). Then \(\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ \mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right|\) is equal to

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Three Dimensional Geometry

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

1 α a 0 β b 1 γ c

is equal to

Answer: (A)

Write points on

L_{1}

(2 p + 1, 3 p + 2, 4 p + 3)

and on

L_{2} : x - 6 = y = - z + 4

(q + 6, q, 4 - q)

, with

P (4, 1, 0)

. Then

P A = (2 p - 3, 3 p + 1, 4 p + 3)

and

P B = (q + 2, q - 1, 4 - q)

. Collinearity of

P, A, B

makes these proportional; solving

\frac{2 p - 3}{q + 2} = \frac{3 p + 1}{q - 1} = \frac{4 p + 3}{4 - q}

gives

p = - 1, q = 3

, so

A (- 1, - 1, - 1)

and

B (9, 3, 1)

. Then

1 - 1 9 0 - 1 3 1 - 1 1 = 1 (- 1 + 3) - 0 + 1 (- 3 + 9) = 2 + 6 = 8.

The value of the determinant is

8

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option