If 2 vec{a} + 3 vec{b} + vec{c} = vec{0},then | vedclass

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(B) Given,$2 \vec{a} + 3 \vec{b} + \vec{c} = \vec{0}$.From this,we have $\vec{c} = -2 \vec{a} - 3 \vec{b}$.Now,substitute $\vec{c}$ into the expressio

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$3(\vec{b} 	imes \vec{c})$

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(B) Given,$2 \vec{a} + 3 \vec{b} + \vec{c} = \vec{0}$.From this,we have $\vec{c} = -2 \vec{a} - 3 \vec{b}$.Now,substitute $\vec{c}$ into the expression $\vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{a}$:$= \vec{a} 	imes \vec{b} + \vec{b} 	imes (-2 \vec{a} - 3 \vec{b}) + (-2 \vec{a} - 3 \vec{b}) 	imes \vec{a}$$= \vec{a} 	imes \vec{b} - 2(\vec{b} 	imes \vec{a}) - 3(\vec{b} 	imes \vec{b}) - 2(\vec{a} 	imes \vec{a}) - 3(\vec{b} 	imes \vec{a})$Since $\vec{x} 	imes \vec{x} = \vec{0}$ and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$:$= \vec{a} 	imes \vec{b} - 2(-\vec{a} 	imes \vec{b}) - 3(\vec{0}) - 2(\vec{0}) - 3(-\vec{a} 	imes \vec{b})$$= \vec{a} 	imes \vec{b} + 2(\vec{a} 	imes \vec{b}) + 3(\vec{a} 	imes \vec{b})$$= 6(\vec{a} 	imes \vec{b})$.Alternatively,using the cross product with $\vec{a}$ and $\vec{b}$:$2(\vec{a} 	imes \vec{a}) + 3(\vec{a} 	imes \vec{b}) + (\vec{a} 	imes \vec{c}) = \vec{0} \Rightarrow 3(\vec{a} 	imes \vec{b}) = \vec{c} 	imes \vec{a}$.$2(\vec{b} 	imes \vec{a}) + 3(\vec{b} 	imes \vec{b}) + (\vec{b} 	imes \vec{c}) = \vec{0} \Rightarrow 2(\vec{a} 	imes \vec{b}) = \vec{b} 	imes \vec{c}$.Substituting these into the expression:$\vec{a} 	imes \vec{b} + 2(\vec{a} 	imes \vec{b}) + 3(\vec{a} 	imes \vec{b}) = 6(\vec{a} 	imes \vec{b})$.Note: The provided options suggest a relation to $\vec{b} 	imes \vec{c}$. Since $2(\vec{a} 	imes \vec{b}) = \vec{b} 	imes \vec{c}$,then $\vec{a} 	imes \vec{b} = \frac{1}{2}(\vec{b} 	imes \vec{c})$.Thus,$6(\vec{a} 	imes \vec{b}) = 6 	imes \frac{1}{2}(\vec{b} 	imes \vec{c}) = 3(\vec{b} 	imes \vec{c})$.

If $2 \vec{a} + 3 \vec{b} + \vec{c} = \vec{0}$,then $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is equal to

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