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B
−[abc]
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C
3[abc]
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D
−3[abc]
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Solution
The correct option is D3[abc] (a+2b−c)⋅((a−b)×(a−b−c)) =(a+2b−c)⋅((a−b)×(a−b)−(a−b)×c) =−(a+2b−c)⋅((a−b)×c) =−(a−b+3b−c)⋅((a−b)×c) =−(3b−c)⋅((a−b)×c)=−3b⋅(a×c)=−3[bac]=3[abc]