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If $\bar{a}, \bar{b}, \bar{c}$ are non $-$ coplanar vectos and $(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}}+\overline{\mathrm{b}} \times \overline{\mathrm{c}}+\overline{\mathrm{c}} \times \overline{\mathrm{a}})=\mathrm{k}[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]$,
then value of $\mathrm{k}$ is
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then value of $\mathrm{k}$ is
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Verified Answer
The correct answer is:
3
$\begin{aligned} &(\bar{a}+\bar{b}+\bar{c}) \cdot[(\bar{a} \times \bar{b})+(\mathrm{b} \times \bar{c})+(\bar{c} \times \bar{a})] \\=&[\bar{a} \cdot(\bar{a} \times \bar{b})]+[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{a} \times \bar{b})]+[\bar{b} \cdot(\bar{b} \times \bar{c})] \\ &+[\bar{b} \cdot(\bar{c} \times \bar{a})]+[\bar{c} \cdot(\bar{a} \times \bar{b})]+[\bar{c} \cdot(\bar{b} \times \bar{c})]+[\bar{c} \cdot(\bar{c} \times \bar{a})] \\=& 0+[\bar{a} \cdot(\bar{b} \times \bar{c})]+0+0+0+[\bar{b} \cdot(\bar{c} \times \bar{a})]+[\bar{c} \cdot(\bar{a} \times \bar{b})]+0+0 \\=& 3[\bar{a} \cdot(\bar{b} \times \bar{c})]=3\left[\begin{array}{lll}a & \bar{b} & \bar{c}\end{array}\right] \Rightarrow \mathrm{k}=3 \end{aligned}$
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