Given α→=3i^+j^+2k^ and β→=i^−2j^−4k^ are the  position vectors of the points A and B. Then the distance   of the point −i^+j^+k^ from the plane passing through B and perpendicular to AB is

AB→=β→−α→=−2i^−3j^−6k^Equation of the plane passing through B and perpendicular to AB is  (r→−OB→)⋅AB→=0r→⋅(2i^+3j^+6k^)+28=0Hence, the required distance from r→=−i^+j^+k^=(−i^+j^+k^)⋅(2i^+3j^+6k^)+28|2i^+3j^+6k^|=−2+3+6+287=5 units