If x, y, z are non-zero real numbers, $\overline{\mathrm{a}}=\mathrm{x}\overline{\mathrm{i}}+2\overline{\mathrm{j}}, \overline{b}=\mathrm{y}\overline{\mathrm{j}}+3\overline{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{x}\overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{j}}$ are such that $\overline{\mathrm{a}}\times \overline{\mathrm{b}}=\mathrm{z}\overline{\mathrm{i}}-3\overline{\mathrm{j}}+\overline{\mathrm{k}}$ then $[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=$

$\overline{a}\times \overline{b}=\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ \mathrm{x}& 2& 0\\ \mathrm{o}& \mathrm{y}& 3\end{array}\right|=6\hat{\mathrm{i}}-3\mathrm{x}\hat{\mathrm{j}}+\mathrm{xy}\hat{\mathrm{k}}$

$$\begin{gathered} given that \\ \overline{\mathrm{a}}\times \overline{\mathrm{b}}=\mathrm{z}\hat{\mathrm{i}}-3\widehat{\mathrm{j}}+\widehat{\mathrm{k}} \end{gathered}$$

$$\begin{gathered} z=6,x=1,y=1 \\ \overline{c}=\widehat{i}+\widehat{j}+6\widehat{k} \\ \therefore [\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=(\overline{\mathrm{a}}\times \overline{\mathrm{b}})\cdot \mathrm{c}=(6\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+\widehat{\mathrm{k}})(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}) \\ =9 \end{gathered}$$