Find →a.(→b×→c), if →a=2^i+^j+3^k,→b=−^i+2^j+^k and →c=3^i+^j+2^k.
As →a.(→b×→c)=[→a→b→c]=⎡⎢⎣213−121312⎤⎥⎦
So, →a.(→b×→c)=2(4−1)−1(−2−3)+3(−1−6)=−10.