Vector equation of the line 6x−2=3y+1=2z−2  is

The line can be written as x−1/31=y+1/32=z−13  which passes through the point whose position vector is (1/3)i−(1/3)j+k  and is parallel to the vector i+2j+3k  and hence its equation is r=13i−13j+k+λ(i+2j+3k).