Let $\mathrm{L}_{1}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{2}$ and $\mathrm{L}_{2}: \frac{\mathrm{x}+1}{-1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}}{1}$ be two lines. Let $\mathrm{L}_{3}$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$. If $\mathrm{L}_{3}$ intersects $\mathrm{L}_{1}$, then $|5 \alpha-11 \beta-8 \gamma|$ equals :

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Three Dimensional Geometry

Let

L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2}

and

L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z}{1}

be two lines.

Let

L_{3}

be a line passing through the point

(α, β, γ)

and be perpendicular to both

L_{1}

and

L_{2}

. If

L_{3}

intersects

L_{1}

, then

∣5 α - 11 β - 8 γ ∣

equals :

Answer: (C)

The direction ratios of

L_{1}

and

L_{2}

are

(1, - 1, 2)

and

(- 1, 2, 1)

. A line perpendicular to both has direction

d = (1, - 1, 2) \times (- 1, 2, 1) = (- 5, - 3, 1) .

L_{3} : \frac{x - α}{- 5} = \frac{y - β}{- 3} = \frac{z - γ}{1} = t

, meeting

L_{1}

(parametrised as

(1 + s, 2 - s, 1 + 2 s)

) at some

s, t

α - 5 t = 1 + s, β - 3 t = 2 - s, γ + t = 1 + 2 s .

Adding the first two eliminates

s

α + β - 8 t = 3

, so

t = \frac{α + β - 3}{8}

.

Combining the first and third also eliminates

s

2 α - γ - 11 t = 1

, so

t = \frac{2 α - γ - 1}{11}

.

Equating the two expressions for

t

11 (α + β - 3) = 8 (2 α - γ - 1) ⟹ 5 α - 11 β - 8 γ = - 25.

∣5 α - 11 β - 8 γ ∣ = 25

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option