The correct option is D (→a×→b)⋅(→b×→c)(→b×→c)⋅(→a×→c)=−1 if →a+→b+→c=0
For the vectors to form right handed system,
1) they should be non-coplanar
2) they should be perpendicular to each other
For option (a)
→a×(→b×→c)=(→a⋅→c)→b−(→a⋅→b)→c (i)
→b×(→c×→a)=(→b⋅→a)→c−(→b⋅→c)→a (ii)
→c×(→a×→b)=(→c⋅→b)→a−(→c⋅→a)→b (iii)
Adding (1), (2) and (3), we get sum as zero hence
→a×(→b×→c)=−→b(→c×→a)−→c(→a×→b)
They are coplanar so can't form right handed system
For option (b)
(→a×→b)×→c⊥→c
and (→a×→b)×→c⊥→a×→b
Also they will be non coplanar
So, option (b) forms right handed system
for option (c)
→a+→b+→c=0
⇒|→a+→b+→c|2=0
|→a|2+→a⋅→b+→a⋅→c+→b⋅→a +|→b|2+→b⋅→c+→c⋅→a+→c⋅→b +|→c|2=0
⇒|→a|2+→b|2+→c|2+2(→a⋅→b+→b⋅→c+→c⋅→a)=0
Since, |→a|2+→b|2+→c|2>0
(→a⋅→b+→b⋅→c+→c⋅→a)<0
For option (d)
→a+→b+→c=0
(→a+→b+→c)×→b=0
⇒→a×→b=→b×→c (i)
Also, (→a+→b+→c)×→c=0
⇒−→a×→c=→b×→c (ii)
putting values of (i) and (ii) in LHS, we get
(→a×→b)⋅(→b×→c)(→b×→c)⋅(→a×→c)=−1
option (d) is correct.