Place one of the vertices at the origin. Every condition is still satisfied. Now let $OACB$ be the parallelogram-shaped face. Let $\vec{OA} = \vec{u}, \vec{OB} = \vec{v}$ for convenience, and $\vec{u}=(a_1 , a_2 , a_3 ), \vec{v}=(b_1 , b_2 , b_3)$. Since the height (the perpendicular edge) is both perpendicular to $\vec{u}, \vec{v}$, It can be written as $\vec{w} = h(\vec{u} \times \vec{v})$ for some constant $h$. Now from the problem's conditions, we have that $\vec{u}, \vec{v}, \vec{w}$ have integral coordinates, and from the condition that there are no integral coordinates on the edges, the gcd of the three components of each vector $\vec{u}, \vec{v}, \vec{w}$ equals 1. From this we also know that $h = \pm \frac{1}{gcd(a_2 b_3 - a_3 b_2 , a_3 b_1 - a_1 b_3 , a_1 b_2 - a_2 b_1 )}$. The volume can be calculated as the following. $(Volume) = (Parallelogram) \times (height) = |\vec{u} | \vec{v} | \sin \theta |\vec{w}| = |h| |\vec{u} \times \vec{v}|^2$ However, we know that $|\vec{u} \times \vec{v}|^2 $ is sum of three squares. Now let's prove that $|h|=1$. Before going on, let's look at a lemma. For arbitrary integers $a, b$, there exists non-trivial solutions to $ax + by \equiv 0 (mod p)$. It's trivial by Chevalley-Warning. Now assume the contrary. Let $p$ be a prime that divides $a_2 b_3 - a_3 b_2 , a_3 b_1 - a_1 b_3 , a_1 b_2 - a_2 b_1$. Case 1. If one of $a_1 , a_2 , a_3 , b_1 , b_2 , b_3$ is a multiple of $p$, WLOG $p|a_1$, then $p$ also divides $a_2 b_1 , a_3 b_1$. $p|b_1$ because, if not, then $p|a_2 , a_3$, a) if $p$ divides one of $a_2 , a_3 , b_2 , b_3$, WLOG $p|a_2$ $\Longrightarrow p|a_3 b_2$ $\Longrightarrow p|b_2$ (since $gcd(a_1 , a_2 , a_3 )=1$) Now take a nontrivial solution of $a_3 x + b_3 y \equiv 0 (mod p)$, $(x_0 , y_0 )$ such that $0 \leq x_0 , y_0 <p$. Easy to check that $\frac{x_0}{p} \vec{u} + \frac{y_0}{p} \vec{v}$ is an integral point which contradicts the condition that there are no integral points on the faces. b) if $p$ doesn't divide any of $a_2 , a_3 , b_2 , b_3$, Take a nontrivial solution of $a_2 x + b_2 y \equiv 0 ( mod p)$, $(x_0 , y_0 )$, and it's easy to check that $a_3 x_0 + b_3 y_0 \equiv 0 ( mod p)$ the rest is analogous to a). Case 2. There are no multiples of $p$, Take a nontrivial solution of $a_1 x + b_1 y \equiv 0 (mod p)$, and the rest is analogous to previous parts. Thus $|h|=1$ and problem solved!