If the vectors bar{a}, bar{b}, bar{c} are non | vedclass

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(C) The scalar triple product is defined as $[\bar{a} \quad \bar{b} \quad \bar{c}] = \bar{a} \cdot (\bar{b} 	imes \bar{c})$.We need to evaluate the e

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$9$

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(C) The scalar triple product is defined as $[\bar{a} \quad \bar{b} \quad \bar{c}] = \bar{a} \cdot (\bar{b} 	imes \bar{c})$.We need to evaluate the expression $[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]$.Using the properties of the scalar triple product,we can write this as:$[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}] = (\bar{a}+2\bar{b}) \cdot [(\bar{b}+2\bar{c}) 	imes (\bar{c}+2\bar{a})]$.Expanding the cross product:$(\bar{b}+2\bar{c}) 	imes (\bar{c}+2\bar{a}) = (\bar{b} 	imes \bar{c}) + 2(\bar{b} 	imes \bar{a}) + 2(\bar{c} 	imes \bar{c}) + 4(\bar{c} 	imes \bar{a})$.Since $\bar{c} 	imes \bar{c} = 0$,this simplifies to $(\bar{b} 	imes \bar{c}) + 2(\bar{b} 	imes \bar{a}) + 4(\bar{c} 	imes \bar{a})$.Now,take the dot product with $(\bar{a}+2\bar{b})$:$(\bar{a}+2\bar{b}) \cdot [(\bar{b} 	imes \bar{c}) + 2(\bar{b} 	imes \bar{a}) + 4(\bar{c} 	imes \bar{a})]$$= \bar{a} \cdot (\bar{b} 	imes \bar{c}) + 2\bar{a} \cdot (\bar{b} 	imes \bar{a}) + 4\bar{a} \cdot (\bar{c} 	imes \bar{a}) + 2\bar{b} \cdot (\bar{b} 	imes \bar{c}) + 4\bar{b} \cdot (\bar{b} 	imes \bar{a}) + 8\bar{b} \cdot (\bar{c} 	imes \bar{a})$.Using the property that the scalar triple product is zero if any two vectors are identical:$= [\bar{a} \bar{b} \bar{c}] + 0 + 0 + 0 + 0 + 8[\bar{b} \bar{c} \bar{a}]$$= [\bar{a} \bar{b} \bar{c}] + 8[\bar{a} \bar{b} \bar{c}] = 9[\bar{a} \bar{b} \bar{c}]$.Therefore,$\frac{[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]}{[\bar{a} \bar{b} \bar{c}]} = \frac{9[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} = 9$.

If the vectors $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar,then $\frac{[\bar{a}+2\bar{b} \quad \bar{b}+2\bar{c} \quad \bar{c}+2\bar{a}]}{[\bar{a} \quad \bar{b} \quad \bar{c}]}=$

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