Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \mathrm{and} \overrightarrow{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}.$ If $\overrightarrow{\mathrm{c}}$ is a vector such that $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}=3,$then $\overrightarrow{\mathrm{a}}.\left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)$ is equal to:

$\begin{array}{rcl}\overline{\mathrm{a}}\times (\overline{\mathrm{a}}\times \overline{\mathrm{c}})& =& \overline{\mathrm{a}}\times \overline{\mathrm{b}}\\ (\overline{\mathrm{a}}·\overline{\mathrm{c}}) \overline{\mathrm{a}}-|\overline{\mathrm{a}}{|}^2& =& \left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 1& 1\\ 0& 1& -1\end{array}\right|\\ 3\overline{\mathrm{a}}-3\overline{\mathrm{c}}& =& \mathrm{i}(-2)-\mathrm{j}(-1)+\mathrm{k}\left(1\right)\\ 3\overline{\mathrm{c}}& =& 3\mathrm{i}+3\mathrm{j}+3\mathrm{k}-2\mathrm{i}-\mathrm{j}-\mathrm{k}\\ & =& 5\mathrm{i}+2\mathrm{j}+2\mathrm{k}\\ \overline{\mathrm{e}}& =& \frac{1}{3} (5\mathrm{i}+2\mathrm{j}+2\mathrm{k})\end{array}$

$\begin{array}{rcl}[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]& =& \frac{1}{3}\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& -1\\ 5& 2& 2\end{array}\right|=\frac{1}{3}\left(1\right(4)-1(5)+1(-5\left)\right)\\ & =& \frac{1}{3} (4-10)=\raisebox{1ex}{$-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.=-2\end{array}$