The length of the perpendicular drawn from (1, 2,3) to the line  x−63=y−72=z−7−2 is

Let P be the point (1,2,3) and PN be the length of the perpendicular from P on the given line. Coordinates of point N are  (3λ+6,2λ+7−2λ+7) Now PN is perpendicular to the given line or vector 3i→+2j→−2k→3(3λ+6−1)+2(2λ+7−2)−2(−2λ+7−3)=0λ=-1Then, point N is (3, 5, 9) PN= 1