The correct option is B 32
In this question, vector →c is not given, therefore, we cannot apply the formulae of (→a×→b×→c) (vector triple product).
Now, |(→a×→b)×→c|=|→a×→b||→c|sin 30∘ ...(i)Again,|→a×→b|=∣∣
∣
∣∣^i^j^k21−2110∣∣
∣
∣∣=2^i−2^j+^k⇒|→a×→b|=√22+(−2)2+1=√4+4+1=√9=3Since,|→c−→a|=2√2 [given]⇒|→c−→a|2=8⇒(→c−→a).(→c−→a)=8⇒→c.→c−→c.→a−→a.→c+→a.→a=8⇒|→c|2+|→a|2−2|→c|=8⇒|→c|2+9−2|→c|=8⇒|→c|2−2|→c|+1=0⇒(|→c|−1)2=0⇒|→c|=1From,Eq.(i),|(→a×→b)×→c|=(3)(1).(12)=32