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veca xx [(vecr-vecb)xxveca]+vecb xx [(ve...

`veca xx [(vecr-vecb)xxveca]+vecb xx [(vecr-vecc)xxvecb]+vecc xx [(vecr-veca)xxvecc]=0 and veca , vecb and vecc` are unit vector mutually perpendicular to each other then the `vecr` is :

A

`(veca-vecb-vecc)/2`

B

`(veca-vecb+vecc)/2`

C

`(veca+vecb+vecc)/2`

D

`(veca+vecb-vecc)/2`

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The correct Answer is:
C
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