# Solve this following

Question:

Let $\quad \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}-\alpha \hat{\mathrm{k}} \quad$ and $\overrightarrow{\mathrm{c}}=-\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$, where $\alpha$ and $\beta$ are integers. If $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-1$ and $\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=10$, then $(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}$ is equal to__________

Solution:

$\overrightarrow{\mathrm{a}}=(1,-\alpha, \beta)$

$\overrightarrow{\mathrm{b}}=(3, \beta,-\alpha)$

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

$\vec{a} \cdot \vec{b}=-1 \Rightarrow 3-\alpha \beta-\alpha \beta=-1$

$\Rightarrow \alpha \beta=2$

$\begin{array}{ll}1 & 2\end{array}$

$2 \quad 1$

$-1-2$

$-2-1$

$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=10$

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

$\Rightarrow 2 \alpha+\beta+5=0$

$\therefore \alpha=-2 ; \beta=-1$

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

$=1(-1+4)-2(3-4)-1(-6+2)$

$=3+2+4=9$