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If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} $are non-coplanar and $ [\overrightarrow{a}\times\overrightarrow{b},\overrightarrow{b}\times\overrightarrow{c},\overrightarrow{c}\times\overrightarrow{a}]=[\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]$ then$[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$ is

\[\begin{array}{1 1}(1)2&(2)3\\(3)1&(4) 0 \end{array}\]

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$ [\overrightarrow{a}\times\overrightarrow{b},\overrightarrow{b}\times\overrightarrow{c},\overrightarrow{c}\times\overrightarrow{a}]=[\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]$
$[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]^2=2 [\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$
$[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=2$
Hence 1 is the correct answer.
answered May 13, 2014 by meena.p
 

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