Let the two lines be l₁ and l₂.

So, and l₂: 3x – y – 2z + 4 = 0 = 2x + y + z + 1

We need to find the shortest distance between l₁ and l₂.

The equation of a plane containing the line l₂ is given by

(3x – y – 2z + 4) + λ(2x + y + z + 1) = 0

⇒ (3 + 2λ)x + (λ – 1) y + (λ – 2) z + (4 + λ) = 0

Direction ratios of l₁ are 2, 4, 1 and those of the line containing the shortest distance are proportional to 3 + 2λ, λ – 1 and λ – 2.

We know that if two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular to each other, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

⇒ (3 + 2λ)(2) + (λ – 1)(4) + (λ – 2)(1) = 0

⇒ 6 + 4λ + 4λ – 4 + λ – 2 = 0

⇒ 9λ = 0

∴ λ = 0

Thus, the plane containing line l₂ is 3x – y – 2z + 4 = 0.

We have

When α = 0, (x, y, z) = (1, 3, –2)

So, the point (1, 3, –2) lies on the line l₁.

Hence, the shortest distance between the two lines is same as the distance of the perpendicular from (1, 3, –2) on to the plane 3x – y – 2z + 4 = 0.

Recall the length of the perpendicular drawn from (x₁, y₁, z₁) to the plane Ax + By + Cz + D = 0 is given by

Here, (x₁, y₁, z₁) = (1, 3, –2) and (A, B, C, D) = (3, –1, –2, 4)

Thus, the required shortest distance is units.