Let the shortest distance between the lines
L: (x-5)/-2=(y-λ)/0=(z+λ)/1, λ ≥ 0 and
L₁: x+1=y-1=4-z be 2 √6. If (α, β, γ) lies on L,
then which of the following is NOT possible?

Solution

Maths sir · 4 steps

Fix λ from the distance first, then write every point of L with one parameter and test each option.

1. Read both lines. L passes through A(5, λ, -λ) with direction d₁⃗ = (-2, 0, 1). Rewrite L₁ as (x+1)/1=(y-1)/1=(z-4)/-1, so it passes through B(-1, 1, 4) with direction d₂⃗ = (1, 1, -1).
2. Common normal. d₁⃗ × d₂⃗ = (-1, -1, -2), with magnitude √6. Also AB⃗ = (-6, 1-λ, 4+λ), and its dot product with the normal is 6 - (1-λ) - 2(4+λ) = -(λ+3).

   (|AB⃗·(d₁⃗×d₂⃗)|)/(|d₁⃗×d₂⃗|) = (|λ+3|)/√6 = 2√6 ⇒ |λ+3| = 12

3. Points on L. This gives λ = 9 or λ = -15. Since λ ≥ 0, λ = 9. Every point of L is then (α, β, γ) = (5-2t, 9, -9+t).

   Option	Expression in t	Possible?
   (a) α+2γ=24	5-2t-18+2t = -13 for every t	Never: always -13
   (b) 2α+γ=7	1-3t = 7	Yes, t=-2
   (c) α-2γ=19	23-4t = 19	Yes, t=1
   (d) 2α-γ=9	19-5t = 9	Yes, t=2

4. So. On L, α + 2γ is stuck at -13, so it can never equal 24. The answer is (a).

The idea to keep. Once a line's points are written with one parameter, any linear expression in the coordinates either takes every value or is a constant. A constant one is the impossible option.

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