If →a=2^i+3^j+6^k,→b=3^i−6^j+2^k,→c=6^i+2^j−3^k, then →a×→b=
Let →a=17(2^i+3^j+6^k),→b=17(6^i+2^j−3^k),→c=c1^i+c2^j+c−3^k and matrix A=⎡⎢ ⎢ ⎢⎣2737676727−37c1c2c3⎤⎥ ⎥ ⎥⎦ If AAT=I, then →c is equal to