The cartesian equation of the line through the points (4,3,5) and which is parallel to the vector 2 ̂i + 5 ̂j + 2 ̂k is-

(x-4)/2 = (y-3)/5 = (z-5)/2

We have vector  a = 4 ̂i + 3 ̂j + 5 ̂k  

and vector b  = 2 ̂i + 5 ̂j + 2 ̂k

Therefore vector equation of the line is given by-

vector r = a+λb

      ⇒ r = (4 ̂i + 3 ̂j + 5 ̂k)+λ (2 ̂i + 5 ̂j + 2 ̂k)

      ⇒ r = (4+ 2λ) ̂i + (3+5λ) ̂j + (5+ 2λ) ̂k

      ⇒  x ̂i + y ̂j +z  ̂k = (4+ 2λ) ̂i + (3+5λ) ̂j + (5+ 2λ) ̂k

Eliminating λ, we get 

(x-4)/2 = (y-3)/5 = (z-5)/2