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(B) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-zero coplanar vectors,their scalar triple product is zero,i.e.,$[\

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(B) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-zero coplanar vectors,their scalar triple product is zero,i.e.,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$. We need to evaluate the scalar triple product $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}]$. Using the definition of the scalar triple product: $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}] = (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [(3 \overrightarrow{b}-\overrightarrow{c}) 	imes (4 \overrightarrow{c}-\overrightarrow{a})]$ $= (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [12(\overrightarrow{b} 	imes \overrightarrow{c}) - 3(\overrightarrow{b} 	imes \overrightarrow{a}) - 4(\overrightarrow{c} 	imes \overrightarrow{c}) + (\overrightarrow{c} 	imes \overrightarrow{a})]$ Since $\overrightarrow{c} 	imes \overrightarrow{c} = 0$,the expression simplifies to: $= (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [12(\overrightarrow{b} 	imes \overrightarrow{c}) - 3(\overrightarrow{b} 	imes \overrightarrow{a}) + (\overrightarrow{c} 	imes \overrightarrow{a})]$ $= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - 6[\overrightarrow{a} \overrightarrow{b} \overrightarrow{a}] + 2[\overrightarrow{a} \overrightarrow{c} \overrightarrow{a}] - 12[\overrightarrow{b} \overrightarrow{b} \overrightarrow{c}] + 3[\overrightarrow{b} \overrightarrow{b} \overrightarrow{a}] - [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$ Since any scalar triple product with two identical vectors is zero,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{a}] = 0, [\overrightarrow{a} \overrightarrow{c} \overrightarrow{a}] = 0, [\overrightarrow{b} \overrightarrow{b} \overrightarrow{c}] = 0, [\overrightarrow{b} \overrightarrow{b} \overrightarrow{a}] = 0$. Thus,the expression becomes: $= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$ Since $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$,we have: $= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 23[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$ Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are coplanar,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$. Therefore,$23 	imes 0 = 0$.

If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are non-zero coplanar vectors,then $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}]$ is

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