The shortest distance between the lines
(x - 3)/2 = (y - 2)/3 = (z - 1)/(- 1) and (x + 3)/2 = (y - 6)/1 = (z - 5)/3, is :

Solution

Maths sir · 4 steps

Project any vector joining the two lines onto d₁⃗ × d₂⃗, the common perpendicular.

1. Read off the lines. Line 1 passes through A(3, 2, 1) with direction d₁⃗ = (2, 3, -1). Line 2 passes through B(-3, 6, 5) with direction d₂⃗ = (2, 1, 3).

   d = (| AB⃗ · (d₁⃗ × d₂⃗) |)/(| d₁⃗ × d₂⃗ |)

2. Common perpendicular. d₁⃗ × d₂⃗ = (3 · 3 - (-1)(1), -(2 · 3 - (-1)(2)), 2 · 1 - 3 · 2) = (10, -8, -4). Its length is √(100 + 64 + 16) = √180 = 6√5.
3. Project AB. AB⃗ = (-3 - 3, 6 - 2, 5 - 1) = (-6, 4, 4). Then AB⃗ · (10, -8, -4) = -60 - 32 - 16 = -108, and its absolute value is 108.
4. So. d = 108/6√5 = 18/√5, which is option (a).

The idea to keep. For two skew lines, the cross product of their directions points along the shortest gap. Project the vector joining any two points of the lines onto it, and that length is the distance.

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