The adjacent sides of a parallelogram are $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$


$\therefore \quad$ Its diagonal is $\vec{a}+\vec{b}$
$=(2 \hat{i}-4 \hat{j}+5 \hat{k})+(\hat{i}-2 \hat{j}-3 \hat{k})=3 \hat{i}-6 \hat{j}+2 \hat{k}$
$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|=7$
$\therefore \quad$ unit vector parallel to diagonal of the parallelgram is $\frac{1}{7}(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ or $\frac{3}{7} \hat{\mathrm{i}}-\frac{6}{7} \hat{\mathrm{j}}+\frac{2}{7} \hat{\mathrm{k}}$
Now $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3\end{array}\right|=22 \hat{i}+11 \hat{j}$
$\therefore \quad$ Area of parallelogram $=|\vec{a} \times \vec{b}|=11 \sqrt{5}$ sq. units. Thus, the unit vector parallel to its diagonal is $\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})$ and area of parallelogram $=11 \sqrt{5}$ sq units.