Let overline{a} = hat{i} + hat{j} + hat{k},ov | vedclass

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(D) Since the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,the scalar triple product $[\overline{a} \, \overline{b} \,

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$-2$

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(D) Since the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,the scalar triple product $[\overline{a} \, \overline{b} \, \overline{c}]$ must be equal to $0$.This is given by the determinant:$\begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & 2 \ x & x-2 & -1 \end{vmatrix} = 0$Expanding the determinant along the first row:$1((-1)(-1) - (2)(x-2)) - 1((1)(-1) - (2)(x)) + 1((1)(x-2) - (-1)(x)) = 0$$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$$5 - 2x + 1 + 2x + 2x - 2 = 0$$2x + 4 = 0$$2x = -4$$x = -2$

Let $\overline{a} = \hat{i} + \hat{j} + \hat{k}$,$\overline{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overline{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$. If the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,then $x = \dots$

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