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Question

For non-coplanar vectors A,B and C, |[ABC]|=|A||B||C| holds if and only if

A
AB=BC=CA=0
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B
AB=0=BC
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C
AB=0=CA
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D
BC=0=CA
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Solution

The correct options are
A AB=BC=CA=0
B AB=0=BC
C AB=0=CA
D BC=0=CA
Let θ be the angle between A and B and ϕ the angle between C
and A×B i.e. the angle between C and the perpendicular to the plane containing A and B. Then
|(A×B)C|=|A×B||C|cosϕ
=|A||B||C|sinθcosϕ
so if the given relation holds, we have sinθcosϕ=1 since
|sinθ||cosϕ|1 so we must have sinθ=1,cosϕ=1 i.e.
θ=π/2,ϕ=0. The former implies that A and B are
perpendicular so that AB=0 On the hand, if ϕ is zero, C must be perpendicular to both A and B
so that BC=0=AC.

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