If the vectors a^i+^j+^k,^i+b^j+^k and ^i+^j+c^k are coplanar (a≠b≠c≠1), then the value of abc−(a+b+c)=
A
2
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B
−2
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C
0
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D
−1
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Solution
The correct option is B−2 The vectors are coplanar. Therefore, their scalar triple product is zero. ⇒∣∣
∣∣a111b111c∣∣
∣∣=0 ⇒a(bc−1)−(c−1)+(1−b)=0 ⇒abc−(a+b+c)=−2