If A→=2i^+3j^+6k^  andB→=3i^−6j^+2k^ , then vector perpendicular to both A→  and B→  has magnitude k times that of(6i^+2j^−3k^) . Then k is equal to

Let C→  be a vector perpendicular to A→  and B→ Then as per question kC→=A→×B→or   k=(A→×B→)C→          =(2i^+3j^+6k^)×(3i^−6j^+2k^)(6i^+2j^−3k^)         =(42i^+14j^−21k^)(6i^+2j^−3k^)=7∵i^×i^=j^×j^=k^×k^=0 i^×j^=k^     j^×k^=i^   k^×i^=j^