A) \[0\]
B) \[1\]
C) \[-1\]
D) None
Correct Answer: C
Solution :
\[=(1+abc)(a-b)(b-c)(c-a)=0\] .... (1) also, , since \[A,\,\,B,\,\,C\] are not coplanar\[=(a-b)(b-c)(c-a)\ne 0\] ... (2) From (1) & (2),\[abc=-1\]You need to login to perform this action.
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