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