Step 1: Let the adjacent sides be $\vec{a} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $\vec{b} = \hat{i} - 2 \hat{j} - 3 \hat{k}$.

Step 2: Finding the diagonal vector: The diagonal of a parallelogram is given by $\vec{a} + \vec{b}$.

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

Step 4: Finding the unit vector: Calculate |$\vec{d}$| = $\sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.

Step 5: The unit vector parallel to the diagonal is $\hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{7} = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$.

Step 6: Finding the area: The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by |$\vec{a} \times \vec{b}$|.

Step 7: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} = \hat{i}$(12+10) - $\hat{j}$(-6-5) + $\hat{k}$(-4+4) = 22$\hat{i} + 11 \hat{j}$.

Step 8: |$\vec{a} \times \vec{b}$| = $\sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605} = 11 \sqrt{5}$.

Step 9: Therefore, the unit vector parallel to the diagonal is $\frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$, and the area is 11$\sqrt{5}$ square units.