Since $\triangle EHG$ is a right triangle, it must have the following side lengths to satisfy the condition that $EH, HG,$ and $EG$ are consecutive integers: $$EH=3\qquad HG=4\qquad EG=5$$Therefore, the dimensions of the base of the parallelepiped are $3$ by $4$, and we already know that it has a height of $12$, so we can find the area: $$A=3\cdot 4\cdot 12=\boxed{144}$$

Since $EH, HG, EG$ are consecutive, we have $EH=3, HG=4, EG=5$. Our answer is hence $3 \cdot 4 \cdot 12 = \boxed{144}$.