We have,
→a=^i+^j+^k,→b=2^i−^j+3^k and →c=^i−2^j+^k
2→a−→b+3→c=2(^i+^j+^k)−(2^i−^j+3^k+3(^i−2^j+^k
=2^i+2^j+2^k−2^i+^j−3^k+3^i−6^j+3^k
=3^i−^j+2^k
∣∣2→a−→b+3→c∣∣=√32+(−3)2+22=√9+9+4=√22
Hence, the unit vector along 2→a−→b+3→c is
2→a−→b+3→c|2→a−→b+3→c|=3^i−3^j+2^k√22=3√22^i−3√22^j+2√22^k.