The correct option is D (→a×→b)+(→b×→c)+(→c×→a)[→a →b →c]
Since →a,→b,→c are three non coplanar vectors, therefore (→a×→b),(→b×→c),(→c×→a) are also non coplanar vectors.
Let →r=x(→a×→b)+y(→b×→c)+z(→c×→a)
Then →r.→a=1⇒1=y[→a →b →c]
⇒y=1[→a →b →c]
Similarly, →r.→b=1⇒1=z[→a →b →c]
⇒z=1[→a →b →c]
and →r.→c=1⇒1=x[→a →b →c]
⇒x=1[→a →b →c]