I'm thinking about a problem: given a cone, generated by some integral vectors $b_1,\cdots,b_n$, which means components of these vectors are integers. And the cone in $\mathbb{R}^n$ is $C:\{\alpha_1b_1+\cdots+\alpha_nb_n|\alpha_i\ge0,\alpha_i\in\mathbb{R},i=1,\cdots,n\}$. We may assume $b_1,\cdots,b_n$ have rank $n$. Consider a hyperplane $H:c^Tx=1$, where $c$ is an integral vector with relatively prime integer components. If the intersectionof $C$ and $H$ is not empty, and assume there is a rational point $v$ in the intersection. And suppose $c^Tx=0$ is a facet of $C$.
I want to know whether there is an integer point in the intersection. Because intuitively, it seems that going out from $v$ by some direction, we are still in the intersection and then we could find an integer point. But I'm confusing about how to show it strictly.
Since $c^Tx=0$ supports a facet of your simplex, it is going to be generated by $n-1$ of the vectors $b_i$, say they are the first $n-1$, so that $c^Tb_i=0$ for all $i=1,\dots, n-1$. The fact that $H\cap C$ is not empty tells you that $c^Tb_n>0$ (take any point $r\in H\cap C$, so $r=\sum_{i=1}^n e_ib_i$ for real numbers $e_i\geq 0$ for all $i$, and $c^Tr=\sum_{i=1}^nc^Te_ib_i= e_n(c^Tb_n)=1$, so $c^Tb_n>0$).
Since the components of $c$ are relatively prime, there is an integral point $\overline{x}$ on the hyperplane $c^Tx=1$ (Bezout's identity). Write $\overline{x}=\sum_{i=1}^n \gamma_i b_i$, where $\gamma_i$ are rational numbers. Since $c^T\overline{x}=1$, you get $\gamma_n(c^Tb_n)=1$, so $\gamma_n\geq 0$.
Now every point of the form $\overline{x}+\sum_{i=1}^{n-1} d_i b_i=\sum_{i=1}^{n-1} (\gamma_i+d_i)b_i +\gamma_n b_n$ where $d_i$ are integers will still be a point with integral coordinates on $H$ (since $c^Tb_i=0$ for $i=1,\cdots, n-1$), and if you take each $d_i$ to be positive and big enough, the corresponding linear combination will also be in $C$.