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(B) We need to evaluate the scalar triple product: $(\overline{a}+\overline{b}+\overline{c}) \cdot [(\overline{a}+\overline{b}) 	imes (\overline{a}+\

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$[\overline{a} \overline{b} \overline{c}]$

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(B) We need to evaluate the scalar triple product: $(\overline{a}+\overline{b}+\overline{c}) \cdot [(\overline{a}+\overline{b}) 	imes (\overline{a}+\overline{c})]$.First,expand the cross product: $(\overline{a}+\overline{b}) 	imes (\overline{a}+\overline{c}) = \overline{a} 	imes \overline{a} + \overline{a} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{b} 	imes \overline{c} = 0 + \overline{a} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{b} 	imes \overline{c}$.Now,take the dot product with $(\overline{a}+\overline{b}+\overline{c})$:$(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{a} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{b} 	imes \overline{c})$.Using the property of scalar triple products $[\overline{x} \overline{y} \overline{z}] = \overline{x} \cdot (\overline{y} 	imes \overline{z})$:$= [\overline{a} \overline{a} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{c} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] + [\overline{c} \overline{b} \overline{c}]$.Since any scalar triple product with two identical vectors is $0$:$= 0 + 0 + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + 0 + 0 + 0 + [\overline{c} \overline{b} \overline{a}] + 0$.Using the property $[\overline{x} \overline{y} \overline{z}] = -[\overline{y} \overline{x} \overline{z}]$:$= [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = [\overline{a} \overline{b} \overline{c}]$.Wait,re-evaluating the expansion: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{a} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{b} 	imes \overline{c}) = [\overline{a} \overline{a} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{c} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] + [\overline{c} \overline{b} \overline{c}] = 0 + 0 + [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + 0 + 0 + 0 + [\overline{a} \overline{b} \overline{c}] + 0 = [\overline{a} \overline{b} \overline{c}]$.Actually,the correct result is $[\overline{a} \overline{b} \overline{c}]$. Let's re-check the expansion: $(\overline{a}+\overline{b}+\overline{c}) \cdot (\overline{a} 	imes \overline{c} + \overline{b} 	imes \overline{a} + \overline{b} 	imes \overline{c}) = [\overline{a} \overline{a} \overline{c}] + [\overline{a} \overline{b} \overline{a}] + [\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{b} \overline{b} \overline{a}] + [\overline{b} \overline{b} \overline{c}] + [\overline{c} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] + [\overline{c} \overline{b} \overline{c}] = 0 + 0 + [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + 0 + 0 + 0 + [\overline{a} \overline{b} \overline{c}] + 0 = [\overline{a} \overline{b} \overline{c}]$.Correction: The original provided solution was $-[\overline{a} \overline{b} \overline{c}]$,but the calculation yields $[\overline{a} \overline{b} \overline{c}]$. Re-calculating: $[\overline{a} \overline{b} \overline{c}] + [\overline{b} \overline{a} \overline{c}] + [\overline{c} \overline{b} \overline{a}] = [\overline{a} \overline{b} \overline{c}] - [\overline{a} \overline{b} \overline{c}] + [\overline{a} \overline{b} \overline{c}] = [\overline{a} \overline{b} \overline{c}]$.

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