Let the values of λ for which the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4
and (x-λ)/3 = (y-4)/4 = (z-5)/5 is 1/√6 be λ₁ and λ₂. Then the radius of the circle passing through the
points (0, 0), (λ₁, λ₂) and (λ₂, λ₁) is

Solution

Maths sir · 4 steps

Cross the direction vectors for the common normal, and the distance condition gives both λ.

1. Common normal. The direction vectors are d₁⃗ = (2, 3, 4) and d₂⃗ = (3, 4, 5). Their cross product is n⃗ = (-1, 2, -1), so |n⃗| = √6.

   d = (|AB⃗ · n⃗|)/|n⃗| = (|-(λ - 1) + 4 - 2|)/√6 = (|3 - λ|)/√6

2. Find both values. Here A = (1, 2, 3) and B = (λ, 4, 5), so AB⃗ = (λ - 1, 2, 2). Setting (|3 - λ|)/√6 = 1/√6 gives |λ - 3| = 1, so λ₁ = 2 and λ₂ = 4.
3. Circle through points. A circle through the origin is x² + y² + 2gx + 2fy = 0. Putting in (2, 4) gives 20 + 4g + 8f = 0, and putting in (4, 2) gives 20 + 8g + 4f = 0. Subtracting the two gives g = f, and then 20 + 12g = 0 gives g = f = -5/3.
4. So. The centre is (5/3, 5/3) and the radius is √(25/9 + 25/9) = 5√2/3, which is option (b).

The idea to keep. For skew lines, the shortest distance is the projection of the joining vector on d₁⃗ × d₂⃗. An absolute value in the condition usually gives two parameter values.

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