If bar(a)=bar(i)+bar(j)+bar(k), bar(b)=4...

If `bar(a)=bar(i)+bar(j)+bar(k), bar(b)=4bar(i)+3bar(j)+4bar(k), bar(c)=bar(i)+alphabar(j)+betabar(k)` are linearly dependent vectors and `abs(bar(c))=sqrt(3)` then

A

`alpha=1, beta=-1`

B

`alpha=1, beta=pm1`

C

`alpha=-1, beta=pm1`

D

`alpha=pm1, beta=1`

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The correct Answer is:

D

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