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(A) Since $\overline{a}, \overline{b}, \overline{c}$ are coplanar vectors,their scalar triple product is zero,i.e.,$[\overline{a}, \overline{b}, \over

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(A) Since $\overline{a}, \overline{b}, \overline{c}$ are coplanar vectors,their scalar triple product is zero,i.e.,$[\overline{a}, \overline{b}, \overline{c}] = 0$. Let $\overline{\alpha} = 2 \overline{a} - \overline{b}$,$\overline{\beta} = 2 \overline{b} - \overline{c}$,and $\overline{\gamma} = 2 \overline{c} - \overline{a}$. The scalar triple product $[\overline{\alpha}, \overline{\beta}, \overline{\gamma}]$ can be expressed as the determinant of the coefficients multiplied by $[\overline{a}, \overline{b}, \overline{c}]$: $[\overline{\alpha}, \overline{\beta}, \overline{\gamma}] = \begin{vmatrix} 2 & -1 & 0 \ 0 & 2 & -1 \ -1 & 0 & 2 \end{vmatrix} [\overline{a}, \overline{b}, \overline{c}]$. Calculating the determinant: $2(4 - 0) - (-1)(0 - 1) + 0 = 2(4) + 1(-1) = 8 - 1 = 7$. Thus,$[\overline{\alpha}, \overline{\beta}, \overline{\gamma}] = 7 	imes [\overline{a}, \overline{b}, \overline{c}] = 7 	imes 0 = 0$.

If $\overline{a}, \overline{b}$ and $\overline{c}$ are unit coplanar vectors,then the scalar triple product $[2 \overline{a}-\overline{b}, 2 \overline{b}-\overline{c}, 2 \overline{c}-\overline{a}]$ has the value

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