If $\quad \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=-\beta \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and
$\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
such that $\vec{a} \cdot \vec{b}=1$ and $\vec{b} \cdot \vec{c}=-3$, then
$\frac{1}{3}((\vec{a} \times \vec{b}) \cdot \vec{c})$ is equal to
$\vec{a} \cdot \vec{b}=1 \Rightarrow-\alpha \beta-\alpha \beta-3=1$
$\Rightarrow-2 \alpha \beta=4 \Rightarrow \alpha \beta=-2$..(1)
$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-3 \quad \Rightarrow \quad-\beta+2 \alpha+1=-3$
$\beta-2 \alpha=4$..(2)
Solving (1) \& (2), $(\alpha, \beta)=(-1,2)$
$\frac{1}{3}[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=\frac{1}{3}\left|\begin{array}{ccc}\alpha & \beta & 3 \\ -\beta & -\alpha & -1 \\ 1 & -2 & -1\end{array}\right|$
$=\frac{1}{3}\left|\begin{array}{ccc}-1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right|$
$=\frac{1}{3}\left|\begin{array}{ccc}0 & 0 & 2 \\ -2 & 1 & -1 \\ 1 & -2 & -1\end{array}\right|=\frac{1}{3}[2(4-1)]=2$
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