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In triangle ABD, we have $\vec{AD} + \vec{DB} = \vec{AB}$. Therefore, $\vec{AD} = \vec{AB} - \vec{DB}$.
Given $\vec{AB}=2\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{DB}=3\hat{i}-6\hat{j}+2\hat{k}$, we can find $\vec{AD}$ as follows: $$ \vec{AD} = (2\hat{i}-4\hat{j}+5\hat{k}) - (3\hat{i}-6\hat{j}+2\hat{k}) = (2-3)\hat{i} + (-4+6)\hat{j} + (5-2)\hat{k} = -\hat{i} + 2\hat{j} + 3\hat{k} $$
The area of parallelogram ABCD is given by the magnitude of the cross product of $\vec{AB}$ and $\vec{AD}$. $$ \text{Area} = |\vec{AB} \times \vec{AD}| $$
$$ \vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ -1 & 2 & 3 \end{vmatrix} = \hat{i}((-4)(3) - (5)(2)) - \hat{j}((2)(3) - (5)(-1)) + \hat{k}((2)(2) - (-4)(-1)) $$ $$ = \hat{i}(-12 - 10) - \hat{j}(6 + 5) + \hat{k}(4 - 4) = -22\hat{i} - 11\hat{j} + 0\hat{k} $$
$$ |\vec{AB} \times \vec{AD}| = \sqrt{(-22)^2 + (-11)^2 + 0^2} = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5} $$
Final Answer: $\vec{AD} = -\hat{i} + 2\hat{j} + 3\hat{k}$ and Area = $11\sqrt{5}$ square units