The shortest distance between the lines x/2 = y/2 = z/1 and x+2 = y = z lies in the interval:

Writing the equation of Line in Vector Form
Let the first line be L1 and the second line be L2.
For L1: x/2 = y/2 = z/1 = λ
Therefore, x = 2λ, y = 2λ, z = λ
The vector equation of L1 is →r1 = λ(2î + 2ĵ + k̂)
For L2: x + 2 = y = z = μ
Therefore, x = μ - 2, y = μ, z = μ
The vector equation of L2 is →r2 = (-2î) + μ(î + ĵ + k̂)
The shortest distance between two lines is given by:
d = |(→b1 × →b2).→a|/|→b1 × →b2|
where →b1 and →b2 are the direction vectors of the lines and →a is the vector joining any point on L1 to any point on L2.
Here, →b1 = 2î + 2ĵ + k̂
→b2 = î + ĵ + k̂
→a = -2î
→b1 × →b2 = (2î + 2ĵ + k̂) × (î + ĵ + k̂) = î(2 - 1) - ĵ(2 - 1) + k̂(2 - 2) = î - ĵ
|(→b1 × →b2).→a| = |(î - ĵ).(-2î)| = |-2| = 2
|→b1 × →b2| = √(1² + (-1)²) = √2
d = 2/√2 = √2 ≈ 1.414
Therefore, the shortest distance lies in the interval [1, 2).