If $\overline{\mathrm{a}},\overline{\mathrm{b}},\overline{\mathrm{c}}$ form a left handed orthogonal system and $\overline{\mathrm{a}}\cdot \overline{\mathrm{a}}=4,\overline{\mathrm{b}}\cdot \overline{\mathrm{b}}=9,\overline{\mathrm{c}}\cdot \overline{\mathrm{c}}=16$ then $(\overline{\mathrm{a}}\times \overline{\mathrm{b}})\cdot \overline{\mathrm{c}}=$

$$\begin{gathered} {\left(\overrightarrow{a}\times \overrightarrow{b}.\overrightarrow{c}\right)}^2={\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]}^2=\left|\begin{array}{ccc}\overrightarrow{a}.\overrightarrow{a}& \overrightarrow{a}.\overrightarrow{b}& \overrightarrow{a}.\overrightarrow{c}\\ \overrightarrow{b}.\overrightarrow{a}& \overrightarrow{b}.\overrightarrow{b}& \overrightarrow{b}.\overrightarrow{c}\\ \overrightarrow{c}.\overrightarrow{a}& \overrightarrow{c}.\overrightarrow{b}& \overrightarrow{c.}\overrightarrow{c}\end{array}\right| \\ =\left|\begin{array}{ccc}4& 0& 0\\ 0& 9& 0\\ 0& 0& 16\end{array}\right|=4\times 9\times 16=576 \end{gathered}$$
$\begin{array}{r}\left(\right(\overline{\mathrm{a}}\times \overline{\mathrm{b}})\cdot \overline{\mathrm{c}}{)}^2=(24{)}^2\\ \because \overline{\mathrm{a}},\overline{\mathrm{b}},\overline{\mathrm{c}} \mathrm{are} in \mathrm{left} \mathrm{hand}ed system\\ \Rightarrow (\overline{\mathrm{a}}\times \overline{\mathrm{b}})\cdot \overline{\mathrm{c}}=-24\end{array}$