Given :
(3→a−2→b)⋅→c=0⇒3→a⋅→c−2→b⋅→c=0⋯(i),
(3→b−2→c)⋅→a=0⇒3→a⋅→b−2→a⋅→c=0⋯(ii)
and
(3→c−2→a)⋅→b=0⇒3→b⋅→c−2→a⋅→b=0⋯(iii)
Adding (i),(ii) and (iii):
we get, (→a⋅→b+→b⋅→c+→a⋅→c)=0⋯(A)
Also,
|→a+→b+→c|2=|→a|2+|→b|2+|→c|2+2(→a⋅→b+→b⋅→c+→a⋅→c)=4+9+3 [using (A)][∵|→a|=2,|→b|=3,|→c|=√3]=16⇒|→a+→b+→c|=4