The cartesian equation of the line through the points (1,2,3) which is parallel to the vector 4 ̂i + 2 ̂j + 6 ̂k is given by- 

(x-1)/4 = (y-2)/2 = (z-3)/6

We have vector  a =  ̂i + 2 ̂j + 3 ̂k  

and vector b  = 4 ̂i + 2 ̂j + 6 ̂k

Therefore vector equation of the line is given by-

vector r = a+λb

      ⇒ r = ( ̂i + 2 ̂j + 3 ̂k)+λ (4 ̂i + 2 ̂j + 6 ̂k)

      ⇒ r = (1+ 4λ) ̂i + (2+2λ) ̂j + (3+ 6λ) ̂k

      ⇒  x ̂i + y ̂j +z  ̂k = (1+ 4λ) ̂i + (2+2λ) ̂j + (3+ 6λ) ̂k

Eliminating λ, we get 

(x-1)/4 = (y-2)/2 = (z-3)/6