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If veca. vecb = 1, vecb.vec c=2 and vecc...

If `veca. vecb = 1, vecb.vec c=2` and `vecc. veca=3`, then the value of `[vecaxx(vecbxxvecc),vecbxx(veccxxveca),veccxx(vecbxxveca)]` is :

A

0

B

`-6veca.(vecbxxvecc)`

C

`12vecc.(vecaxxvecb)`

D

`-12vecb.(veccxxveca)`

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