Given,

and Position Vectors (Vector from origin to a point is called a position vector)

∴ line passing through A and along AB will have equation

and line passing through C and along CD will have equation

Now, PQ is a vector perpendicular to both AB and CD, such that P lies on AB and Q lies on CD

Therefore, coordinates of P and Q will have the form

P (6 + 3λ, 7 – λ, 4 + λ) (i)

Q (-3μ, 2μ – 9, 2 + 4μ) (ii)

Hence, vector PQ will be

Now, As PQ is perpendicular to both, therefore dot products

AB. PQ = 0 and CD. PQ = 0

AB. PQ = 3(-3μ – 6 – 3λ) – (2μ – 16 + λ) + (4μ – 2 – λ)

⇒ 0 = -9μ – 18 – 9λ – 2μ + 16 - λ + 4μ – 2 - λ

⇒ -7μ – 11λ - 4 = 0 …. (iii)

CD.PQ = 3(-3μ – 6 – 3λ) + 2(2μ – 16 + λ) + 4(4μ – 2 – λ)

⇒ 0 = -9μ – 18 – 9λ + 4μ - 32 + 2λ +16 μ – 8 - 4λ

⇒ 29μ + 7λ - 22 = 0 …. (iv)

Solving (iii) and (iv), we get

λ = -1 and μ = 1

Put value of λ in (i) to get,

P (6 + 3(-1), 7 – (-1), 4 + (-1))

P (6-3,7+1,4-1)

P (3,8,3)

Put value of μ in (ii) to get,

Q (-3(1), 2(1) – 9, 2 + 4(1))

Q (-3, 2 – 9, 2 +4)

Q (3, -7, 6)

Hence, position vector of P and Q will be