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(B) Given $\vec{a}+\vec{b} = \lambda \vec{c} \quad \dots(i)$ and $\vec{b}+\vec{c} = \mu \vec{a} \quad \dots(ii)$.Subtracting $(ii)$ from $(i)$,we get

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$-3$

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(B) Given $\vec{a}+\vec{b} = \lambda \vec{c} \quad \dots(i)$ and $\vec{b}+\vec{c} = \mu \vec{a} \quad \dots(ii)$.Subtracting $(ii)$ from $(i)$,we get $\vec{a}-\vec{c} = \lambda \vec{c} - \mu \vec{a}$.Rearranging terms,$(1+\mu)\vec{a} = (1+\lambda)\vec{c}$.Since two vectors are collinear,let $\vec{a}$ and $\vec{b}$ be collinear,then $\vec{b} = k\vec{a}$.Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}$,we have $|k\vec{a}| = |\vec{a}| \Rightarrow |k|=1$,so $k=1$ or $k=-1$.If $k=1$,$\vec{b}=\vec{a}$,then $\vec{a}+\vec{a} = 2\vec{a}$ is collinear with $\vec{c}$,so $\vec{c} = \pm \vec{a}$.If $\vec{c} = -\vec{a}$,then $\vec{a}+\vec{b}+\vec{c} = \vec{a}+\vec{a}-\vec{a} = \vec{a} 
eq 0$. However,the condition $\vec{a}+\vec{b}+\vec{c}=0$ is derived from the collinearity constraints.Using $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$.Since $\vec{a}+\vec{b}+\vec{c}=0$,we have $0 = 2+2+2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$.Thus,$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = -6$.Therefore,$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = -3$.

Given three vectors $\bar{a}, \bar{b}, \bar{c}$,two of which are collinear. If $\bar{a}+\bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+\bar{c}$ is collinear with $\bar{a}$,and $|\bar{a}|=|\bar{b}|=|\bar{c}|=\sqrt{2}$,then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$

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