Let vec{lambda} = xvec{a} + yvec{b} + zvec{c} | vedclass

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(D) Given $\vec{\lambda} = x\vec{a} + y\vec{b} + z\vec{c}$ and $\vec{\lambda} \cdot (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes

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$2$

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(D) Given $\vec{\lambda} = x\vec{a} + y\vec{b} + z\vec{c}$ and $\vec{\lambda} \cdot (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}) = 2(x + y + z)$.Substituting $\vec{\lambda}$ into the dot product equation:$(x\vec{a} + y\vec{b} + z\vec{c}) \cdot (\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}) = 2(x + y + z)$.Expanding the dot product,we use the property that $\vec{a} \cdot (\vec{a} 	imes \vec{b}) = 0$,$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = [\vec{a} \, \vec{b} \, \vec{c}]$,etc.$= x[\vec{a} \, \vec{b} \, \vec{c}] + x[\vec{a} \, \vec{c} \, \vec{a}] + y[\vec{b} \, \vec{a} \, \vec{b}] + y[\vec{b} \, \vec{b} \, \vec{c}] + y[\vec{b} \, \vec{c} \, \vec{a}] + z[\vec{c} \, \vec{a} \, \vec{b}] + z[\vec{c} \, \vec{b} \, \vec{c}] + z[\vec{c} \, \vec{c} \, \vec{a}]$Since any scalar triple product with two identical vectors is $0$,this simplifies to:$= x[\vec{a} \, \vec{b} \, \vec{c}] + y[\vec{b} \, \vec{c} \, \vec{a}] + z[\vec{c} \, \vec{a} \, \vec{b}] = 2(x + y + z)$.Using the cyclic property $[\vec{a} \, \vec{b} \, \vec{c}] = [\vec{b} \, \vec{c} \, \vec{a}] = [\vec{c} \, \vec{a} \, \vec{b}]$,we get:$(x + y + z)[\vec{a} \, \vec{b} \, \vec{c}] = 2(x + y + z)$.Since $x + y + z 
eq 0$,we divide both sides by $(x + y + z)$ to obtain:$[\vec{a} \, \vec{b} \, \vec{c}] = 2$.

Let $\vec{\lambda} = x\vec{a} + y\vec{b} + z\vec{c}$ and $\vec{\lambda} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = 2(x + y + z)$ (where $x + y + z \neq 0$),then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ is:

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