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Let $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{x} \hat{\mathrm{i}}+(\mathrm{x}-2) \hat{\mathrm{j}}-\hat{\mathrm{k}}$. If the vector $\overline{\mathrm{c}}$ lies in the plane of $\bar{a}$ and $\bar{b}$, then $x$ equals
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The correct answer is:
$-2$
$-2$
$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$
$\left|\begin{array}{ccr}x & x-2 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 2\end{array}\right|=0$
$3 x+2-x+2=0$
$2 x=-4$
$x=-2$.
$\left|\begin{array}{ccr}x & x-2 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 2\end{array}\right|=0$
$3 x+2-x+2=0$
$2 x=-4$
$x=-2$.
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