In the given figure, ABCD is a parallelogram. If $\displaystyle \overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{D B}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then find $\displaystyle \overrightarrow{A D}$ and hence find the area of parallelogram ABCD .

$$\begin{aligned} & \overrightarrow{A D}+\overrightarrow{D B}=\overrightarrow{A B} \\ & \overrightarrow{A D}=(2 \widehat{\imath}-4 \widehat{\jmath}+5 \widehat{\mathrm{k}})-(3 \widehat{\imath}-6 \widehat{\jmath}+2 \widehat{\mathrm{k}}) \\ & =-\widehat{\imath}+2 \widehat{\jmath}+3 \widehat{\mathrm{k}} \\ & \overrightarrow{A D} \times \overrightarrow{A B}=\left|\begin{array}{ccc} \widehat{\imath} & \widehat{\jmath} & \widehat{\mathrm{k}} \\ -1 & 2 & 3 \\ 2 & -4 & 5 \end{array}\right|=22 \widehat{\imath}+11 \widehat{\jmath} \\ & \begin{array}{c} \text { Area }=|\overrightarrow{A D} \times \overrightarrow{A B}|=|22 \widehat{\imath}+11 \widehat{\jmath}| \\ =\sqrt{605} \text { or } 11 \sqrt{5} \end{array} \end{aligned} $$