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(D) For three vectors $\overrightarrow{A}$,$\overrightarrow{B}$,and $\overrightarrow{C}$ to be coplanar,their scalar triple product must be zero,i.e.,

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No value of $C_1$ can be found

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(D) For three vectors $\overrightarrow{A}$,$\overrightarrow{B}$,and $\overrightarrow{C}$ to be coplanar,their scalar triple product must be zero,i.e.,$[\overrightarrow{A} \, \overrightarrow{B} \, \overrightarrow{C}] = 0$. The scalar triple product is given by the determinant: $[\overrightarrow{A} \, \overrightarrow{B} \, \overrightarrow{C}] = \begin{vmatrix} 1 & 1 & 1 \ 1 & 0 & 0 \ C_1 & C_2 & C_3 \end{vmatrix} = 0$. Substituting the given values $C_2 = -1$ and $C_3 = 1$: $[\overrightarrow{A} \, \overrightarrow{B} \, \overrightarrow{C}] = \begin{vmatrix} 1 & 1 & 1 \ 1 & 0 & 0 \ C_1 & -1 & 1 \end{vmatrix} = 0$. Expanding along the second row: $-1 	imes \begin{vmatrix} 1 & 1 \ -1 & 1 \end{vmatrix} = 0 \Rightarrow -1 	imes (1 - (-1)) = 0 \Rightarrow -1 	imes (2) = 0 \Rightarrow -2 = 0$. Since $-2 
eq 0$,the scalar triple product is independent of $C_1$ and is never zero. Therefore,no value of $C_1$ can make these vectors coplanar.

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

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