In the figure, $\mathrm{OAC}$ is a triangle and $\mathrm{OB}$ is a median such that
$\begin{array}{l}
\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\
\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}} \text { (say) }
\end{array}$


$\begin{array}{l}
\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OB}} \\
\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}=2 \overrightarrow{\mathrm{b}} \\
\Rightarrow \overrightarrow{\mathrm{c}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=2(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
=(3 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})
\end{array}$
Now, the area of the triangle,
$\Delta=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|$
Here,
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \mathrm{k} \\ 1 & -1 & 1 \\ 3 & 9 & 5\end{array}\right|$
$=\hat{\mathrm{i}}(-5-9)-\hat{\mathrm{j}}(5-3)+\hat{\mathrm{k}}(9+3)$
$=-14 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+1 \hat{2} \mathrm{k}=2(-7 \hat{\mathrm{i}}-\hat{\mathrm{j}}+6 \hat{\mathrm{k}})$
$\begin{aligned} \therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}| &=2 \sqrt{(-7)^{2}+(-1)^{2}+(6)^{2}} \\ &=2 \sqrt{49+1+36}=2 \sqrt{86} \end{aligned}$
$\therefore \Delta=\frac{1}{2} \times 2 \sqrt{86}=\sqrt{86}$