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If $\quad \bar{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=-\hat{i}+2 \hat{j}-2 \hat{k} \quad$ and $\quad \bar{c}=2 \hat{i}-\hat{j}+2 \hat{k} \quad$ then $(\bar{a}-\bar{b}) \cdot[(\bar{a} \times \bar{b}) \times(\bar{a} \times \bar{c})]$ is
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$\begin{aligned} & (\vec{a}-\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a} \times \vec{c})] \\ & =(\vec{a}-\vec{b}) \cdot\{[\vec{a} \vec{b} \vec{c}] \vec{a}-[\vec{a} \vec{b} \vec{a}] \vec{c}\} \\ & =(\vec{a}-\vec{b}) \cdot \vec{a}\{[\vec{a} \vec{b} \vec{c}]\} \\ & =(\vec{a} \cdot \vec{a}-\vec{b} \cdot \vec{a})[\vec{a} \vec{b} \vec{c}]\end{aligned}$
$\begin{aligned} & =\left\{\left(1^2+1^2+1^2\right)-(-1+2-2)\right\}\left|\begin{array}{ccc}1 & 1 & 1 \\ -1 & 2 & -2 \\ 2 & -1 & 2\end{array}\right| \\ & =4 \times\left|\begin{array}{ccc}1 & 0 & 0 \\ -1 & 3 & -1 \\ 2 & -3 & 0\end{array}\right| \\ & =4 \times 1 \times(0-3)=-12\end{aligned}$
$\begin{aligned} & =\left\{\left(1^2+1^2+1^2\right)-(-1+2-2)\right\}\left|\begin{array}{ccc}1 & 1 & 1 \\ -1 & 2 & -2 \\ 2 & -1 & 2\end{array}\right| \\ & =4 \times\left|\begin{array}{ccc}1 & 0 & 0 \\ -1 & 3 & -1 \\ 2 & -3 & 0\end{array}\right| \\ & =4 \times 1 \times(0-3)=-12\end{aligned}$
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