If a → = α i ^ + β j ^ + 3 k ^ ,   b → = - β i ^ - α j ^ - k ^ and c…

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

$\Rightarrow -2\alpha \beta =4\Rightarrow \boxed{\alpha \beta =-2} \dots \left(1\right)$

$\overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{c}}=-3\Rightarrow -\beta +2\alpha +1=-3$

$\boxed{\beta -2\alpha =4} \dots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right),$ $(\alpha ,\beta )=(-1,2)$

$\frac{1}{3}\left[\overrightarrow{\mathrm{a}}\overrightarrow{\mathrm{b}}\overrightarrow{\mathrm{c}}\right]=\frac{1}{3}\left|\begin{array}{ccc}\alpha & \beta & 3\\ -\beta & -\alpha & -1\\ 1& -2& -1\end{array}\right|$

$=\frac{1}{3}\left|\begin{array}{ccc}-1& 2& 3\\ -2& 1& -1\\ 1& -2& -1\end{array}\right|$

$R_1\to R_1+R_3$

$=\frac{1}{3}\left|\begin{array}{ccc}0& 0& 2\\ -2& 1& -1\\ 1& -2& -1\end{array}\right|=\frac{1}{3}\left[2\left(4-1\right)\right]=2$