If the shortest distance between the lines x/α = (y+1) = z/1, (α ≠ 1) and x+y+z+1=0=2x-y+z+3 is 1/√3, then a value of α is :

Shortest distance between the lines x/α = (y+1) = z/1, (α ≠ 1) and x+y+z+1=0=2x-y+z+3 is 1/√3
Line of intersection of planes x+y+z+1=0=2x-y+z+3 is
Eliminating y gives 3x+2z+4=0 ⇒ x = -2z/3 - 4/3 -(1)
Substituting above x in x+y+z+1=0 gives 3y+z-1=0 ⇒ z = -3y+1 -(2)
From (1) and (2), line equation is 3x+4 = -3y+1 = z ⇒ x - (-4/3) = y - (1/3) = z/1
Given line is x/α = (y+1) = z/1
Shortest distance between above two skew lines is (→b - →a) ⋅ (→c × →d) / |→c × →d|
where one line is passing through →a and parallel to →c and other one passes through →b and parallel to →d
→a = (0, -1, 0) →b = (-4/3, 1/3, 0) →c = (α, 1, 1) →d = (1, 1, 3)
∴ (→b - →a) ⋅ (→c × →d) / |→c × →d| = |(-4/3, 4/3, 0) ⋅ (2α-1, 3α-1, α-1)| / |(2α-1, 3α-1, α-1)| = 1/√3
Expanding determinants and simplifying gives 48α² - 64α + 12 = 10α² + 16α + 12 ⇒ 38α² = 64α ⇒ α = 64/38 = 32/19
Hence, option A.