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B
−1
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C
¯¯¯0
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D
2
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Solution
The correct option is C¯¯¯0 →a×→b=→b×→c (→a+→c)×→b=0 So (→a+→c)∥→b →a+→c=K→b−−−−−1 →b×→c=→c=0 (→b+→a)∥→c →b+→a=K→c−−−−−−−2 Compare 1 & 2 K=−1 or K=−1 So ^b+^a+→c=0