The two adjacent sides of a parallelogram are

Adjacent sides of a parallelogram are given as: $\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}$

Then, the diagonal of a parallelogram is given by $\vec{a}+\vec{b}$.

$\vec{a}+\vec{b}=(2+1) \hat{i}+(-4-2) \hat{j}+(5-3) \hat{k}=3 \hat{i}-6 \hat{j}+2 \hat{k}$

Thus, the unit vector parallel to the diagonal is

$\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{3^{2}+(-6)^{2}+2^{2}}}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{9+36+4}}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{7}=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$

$\therefore$ Area of parallelogram $\mathrm{ABCD}=|\vec{a} \times \vec{b}|$

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3\end{array}\right|$

$=\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)$

$=22 \hat{i}+11 \hat{j}$

$=11(2 \hat{i}+\hat{j})$

$\therefore|\vec{a} \times \vec{b}|=11 \sqrt{2^{2}+1^{2}}=11 \sqrt{5}$

Hence, the area of the parallelogram is $11 \sqrt{5}$ square units.