Let $\overrightarrow{a}=\hat{i}-\alpha \hat{j}+\beta \hat{k},\overrightarrow{b}=3\hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\overrightarrow{c}=-\alpha \hat{i}-2\hat{j}+\hat{k},$ where $\alpha$ and $\beta$ are integers. If $\overrightarrow{a}·\overrightarrow{b}=-1$ and $\overrightarrow{b}·\overrightarrow{c}=10,$ then $\left(\overrightarrow{a}\times \overrightarrow{b}\right)·\overrightarrow{c}$ is equal to ______.

We have, $\overrightarrow{a}=\left(1,-\alpha ,\beta \right),\overrightarrow{b}=\left(3,\beta ,-\alpha \right)$ and

$\overrightarrow{c}=\left(-\alpha ,-2,1\right);\alpha ,\beta \in I$

$\overrightarrow{a}·\overrightarrow{b}=-1\Rightarrow 3-\alpha \beta -\alpha \beta =-1$

$\Rightarrow \alpha \beta =2 ...\left(1\right)$

Now, $\overrightarrow{b}·\overrightarrow{c}=10$

$\Rightarrow -3\alpha -2\beta -\alpha =10$

$\Rightarrow 2\alpha +\beta +5=0 ...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow 2\alpha +\frac{2}{\alpha }+5=0$

$\Rightarrow 2{\alpha }^2+2+5\alpha =0$

$\Rightarrow 2{\alpha }^2+5\alpha +2=0$

$\Rightarrow 2{\alpha }^2+4\alpha +\alpha +2=0$

$\Rightarrow 2\alpha \left(\alpha +2\right)+1\left(\alpha +2\right)=0$

$\Rightarrow \left(2\alpha +1\right)\left(\alpha +2\right)=0$

$\alpha =\frac{-1}{2}$ is neglected since$\alpha$ is an integer.

$\therefore \alpha =-2$

Put $\alpha =-2$ in equation $\left(1\right)$, we get $\beta =-1$.

Now, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\left|\begin{array}{ccc}1& 2& -1\\ 3& -1& 2\\ 2& -2& 1\end{array}\right|$

$=1\left(-1+4\right)-2\left(3-4\right)-1\left(-6+2\right)$

$=3+2+4=9.$