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If $\bar{a}+2 \bar{b}+3 \bar{c}=\overline{0}$ and $(\bar{b} \times \bar{b})+(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})=\lambda(\bar{b} \times \bar{c})$, then $\lambda$ has the value
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Verified Answer
The correct answer is:
6
$\begin{aligned}
& \Rightarrow(\vec{a}+\vec{b}+\vec{c}) \times(\vec{a}+2 \vec{b}+3 \vec{c})=\overrightarrow{0} \\
& \Rightarrow 2 \vec{a} \times \vec{b}+3 \vec{a} \times \vec{c}+\vec{b} \times \vec{a}+3 \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+2 \vec{c} \times \vec{b}=\overrightarrow{0} \\
& \Rightarrow \vec{a} \times \vec{b}+\vec{b} \times \vec{c}-2 \vec{c} \times \vec{a}=\overrightarrow{0} \\
& \Rightarrow \vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=3 \vec{c} \times \vec{a}=3 \vec{c} \times(-2 \vec{b}-3 \vec{c}) \quad[\text { from (1)] } \\
& \Rightarrow \vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=-6 \vec{c} \times \vec{b} \\
& \Rightarrow \vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=6 \vec{b} \times \vec{c} \\
& \Rightarrow \lambda=6
\end{aligned}$
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