Let L₁: (x-1)/1=(y-2)/-1=(z-1)/2 and L₂: (x+1)/-1=(y-2)/2=z/1 be two lines.
Let L₃ be a line passing through the point (α, β, γ) and be perpendicular to both L₁ and L₂. If L₃ intersects L₁, then |5 α-11 β-8 γ| equals :

Solution

Maths sir · 4 steps

Any line perpendicular to both and meeting L₁ lies in one fixed plane through L₁. Find that plane.

1. Direction of L3. L₃ is perpendicular to d₁⃗ = (1, -1, 2) and d₂⃗ = (-1, 2, 1), so it runs along their cross product: d₁⃗ × d₂⃗ = (-5, -3, 1).
2. Plane containing both. L₃ meets L₁, so L₃ and L₁ lie in one plane. That plane contains the point (1, 2, 1) of L₁ and both directions (1, -1, 2) and (-5, -3, 1). Its normal is their cross product.

   (1, -1, 2) × (-5, -3, 1) = (-1+6, -(1+10), -3-5) = (5, -11, -8)

3. Equation of plane. Through (1, 2, 1): 5(x-1) - 11(y-2) - 8(z-1) = 0, which gives 5x - 11y - 8z = -25. Every point of L₃, including (α, β, γ), lies on this plane.
4. So. 5α - 11β - 8γ = -25, so |5α - 11β - 8γ| = 25, option (a).

The idea to keep. When a point is free to move along a line that meets a known line, do not chase parameters. Find the plane holding both lines; the point always satisfies its equation.

Answer from the NTA final answer key. Solution by PrepWiser.