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

The point A(6,7,7) is on the line, let the perpendicular from P meet the line in L. thenAP2=(6−1)2+(7−2)2+(7−3)2=66 Also AL=projection of AP on line (actual d.c’s 317,217,−217 )⇒(6−1).317+(7−2).217+(7−3)−217=17 ∴⊥ distance d of P from the line is given by d2=AP2−AL2=66−17=49 so that d = 7