The correct option is A B,D,C,A
We now,
→a×→b=∣∣
∣
∣∣^i^j^ka1a2a3b1b2b3∣∣
∣
∣∣
For
A.[→a×→b →b×→c →c×→a]
here →a×→b=^i+^j−2^k,
→b×→c=−2^i−5^j+4^k,
→c×→a=−^i+2^j−^k
⇒[→a×→b →b×→c →c×→a]=∣∣
∣∣11−2−2−54−12−1∣∣
∣∣=9
for
B.[→a+→b →b+→c →c+→a]=∣∣
∣∣312324234∣∣
∣∣=−6
for
C.[→a→b→c][→a1 →b1 →c1]=1 and
for
D.[→a−→b →b−→c →c−→a]=∣∣
∣∣−1101−2−2012∣∣
∣∣=0
So, ascending order is : B,D,C,A