The vectors from origin to the points A and B are $\overrightarrow{\mathrm{A}}=3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\mathrm{and}\overrightarrow{\mathrm{B}}=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}$ respectively. The area of the triangle OAB be

Given $\overline{\mathrm{OA}}=\overline{\mathrm{a}}=3\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}\mathrm{and}$

$\overline{\mathrm{OB}}=\overline{\mathrm{b}}=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}-2\widehat{\mathrm{k}}$

$\therefore \left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)=\left(\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 3& -6& 2\\ 2& 1& -2\end{array}\right)$

= $\left(12-2\right)\widehat{\mathrm{i}}+\left(4+6\right)\widehat{\mathrm{j}}+\left(3+12\right)\widehat{\mathrm{k}}$

= $10\widehat{\mathrm{i}}+10\widehat{\mathrm{j}}+15\widehat{\mathrm{k}}\Rightarrow |\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|$

= $\sqrt{{10}^2+{10}^2+{15}^2}$

= $\sqrt{425} = 5\sqrt{17}$

Area of $∆\mathrm{OAB} = =\frac{1}{2}|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\frac{5\sqrt{17}}{2}\mathrm{sq}.\mathrm{unit}.$