→A=2^i+^j,3^j−^k and →C=6^i−2^k. Value of →A−2→B+3→C would be
20^i−5^j+4^k
20^i−5^j−4^k
4^i+5^j+20^k
5^i+4^j+10^k
→A−2→B+3→C=2(^i+^j)−2(3^j−^k)+3(6^i−2^k) =2^i+^j−6^j+2^k+18^i−6^k=20^i−5^j−4^k
→A=(2^i+3^j−^k)
→B=(2^i+5^j−2^k)
find (3→A−2→B)