(vec{a}+2 vec{b}-vec{c}) cdot {(vec{a}-vec{b} | vedclass

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(C) Let the given expression be $E = (\vec{a}+2 \vec{b}-\vec{c}) \cdot \{(\vec{a}-\vec{b}) 	imes (\vec{a}-\vec{b}-\vec{c})\}$.First, simplify the cro

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$3[\vec{a} \vec{b} \vec{c}]$

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(C) Let the given expression be $E = (\vec{a}+2 \vec{b}-\vec{c}) \cdot \{(\vec{a}-\vec{b}) 	imes (\vec{a}-\vec{b}-\vec{c})\}$.First, simplify the cross product term: $(\vec{a}-\vec{b}) 	imes (\vec{a}-\vec{b}-\vec{c}) = (\vec{a}-\vec{b}) 	imes \vec{a} - (\vec{a}-\vec{b}) 	imes \vec{b} - (\vec{a}-\vec{b}) 	imes \vec{c}$.$= (\vec{a} 	imes \vec{a} - \vec{b} 	imes \vec{a}) - (\vec{a} 	imes \vec{b} - \vec{b} 	imes \vec{b}) - (\vec{a} 	imes \vec{c} - \vec{b} 	imes \vec{c})$.Since $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{b} = 0$, this becomes: $0 + \vec{a} 	imes \vec{b} - \vec{a} 	imes \vec{b} + 0 - \vec{a} 	imes \vec{c} + \vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{c} - \vec{a} 	imes \vec{c}$.Now, substitute this back into the expression: $E = (\vec{a}+2 \vec{b}-\vec{c}) \cdot (\vec{b} 	imes \vec{c} - \vec{a} 	imes \vec{c})$.$= \vec{a} \cdot (\vec{b} 	imes \vec{c}) - \vec{a} \cdot (\vec{a} 	imes \vec{c}) + 2 \vec{b} \cdot (\vec{b} 	imes \vec{c}) - 2 \vec{b} \cdot (\vec{a} 	imes \vec{c}) - \vec{c} \cdot (\vec{b} 	imes \vec{c}) + \vec{c} \cdot (\vec{a} 	imes \vec{c})$.Using properties of scalar triple product, terms like $\vec{a} \cdot (\vec{a} 	imes \vec{c}) = 0$ and $\vec{b} \cdot (\vec{b} 	imes \vec{c}) = 0$ and $\vec{c} \cdot (\vec{b} 	imes \vec{c}) = 0$ and $\vec{c} \cdot (\vec{a} 	imes \vec{c}) = 0$.So, $E = [\vec{a} \vec{b} \vec{c}] - 2[\vec{b} \vec{a} \vec{c}] = [\vec{a} \vec{b} \vec{c}] + 2[\vec{a} \vec{b} \vec{c}] = 3[\vec{a} \vec{b} \vec{c}]$.

$(\vec{a}+2 \vec{b}-\vec{c}) \cdot \{(\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c})\} =$

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