Does anyone have a solution to this problem? Thanks!

You have to see this in a different vision. Let's say $ a=-x+y+z, b=x-y+z,c=x+y-z $. Then the planes are just $ a=n,b=n,c=n,a+b+c=n $ which makes the problem a lot more easier.

Does this problem related to lattice theory?

As suggested by toto1234567890, we transform the problem by setting the graph in new variables $$a=-x+y+z,b=x-y+z,c=x+y-z.$$The planes appeared in the graph transform to $a=n,b=n,c=n$ and $a+b+c=n$ for $n\in \mathbb{Z}^+$. Note that in each unit cube, says it's the one with vertices $(t_1,t_2,t_3)$ where $t_1,t_2,t_3\in \{ 0,1\}$, there exists two planes of the form $x+y+z=n$ that cut through the interior of the cube. These two planes cut the cube into three pieces, two tetrahedrons and one octahedron. We also denote $a_0=-x_0+y_0+z_0, b_0=x_0-y_0+z_0$ and $c_0=x_0+y_0-z_0$. Note that we've $a_0,b_0,c_0,a_0+b_0+c_0\not\in \mathbb{Z}$. We need to show that there exists positive integer $k$ that $(ka_0,kb_0,kc_0)$ lies strictly inside the octahedron obtained from some unit cube. Clearly, we must consider the unit cube with vertices $$(\lfloor ka_0\rfloor +t_1,\lfloor kb_0\rfloor +t_2,\lfloor kc_0\rfloor +t_3)$$where $t_1,t_2,t_3\in \{ 0,1\}$. To meet our condition, we need to verify that $$1<(ka_0-\lfloor ka_0\rfloor )+(kb_0-\lfloor kb_0\rfloor )+(kc_0-\lfloor kc_0\rfloor) <2$$and $ka_0,kb_0,kc_0\not\in \mathbb{Z}$. In other words, $1<\{ ka_0\} +\{ kb_0\} +\{ kc_0\} <2$ and $ka_0,kb_0,kc_0\not\in \mathbb{Z}$. Let $a_1=\{ a_0\} ,b_1=\{ b_0\} ,c_1=\{ c_0\}$. We've $a_1,b_1,c_1\in (0,1),a_1+b_1+c_1\not\in \mathbb{Z}$. Also, the condition is equivalent to $1<\{ ka_1\} +\{ kb_1\} +\{ kc_1\} <2$ and $ka_1,kb_1,kc_1\not\in \mathbb{Z}$. If $1<a_1+b_1+c_1<2$, we are done by choosing $k=1$ (since $\{ a_1\} =a_1$ and so do others.) If $a_1+b_1+c_1<1$. Choosing $k=N-1$ where $N$ is a positive integer that $Na_1,Nb_1,Nc_1\in \mathbb{Z}$ (exists since $a_1,b_1,c_1\in \mathbb{Q}$.) We've $\{ ka_1\} =1-a_1$ and so do others. So, $\{ ka_1\} +\{ kb_1\} +\{ kc_1\} =3-(a_1+b_1+c_1)>2$. So, there must exists smallest positive integer $k_0$ that $\{ k_0a_1\} +\{ k_0b_1\} +\{ k_0c_1\} \geq1$. Note that $k_0>1$. We wish to show that $\{ k_0a_1\} +\{ k_0b_1\} +\{ k_0c_1\} <2$ and $k_0a_1,k_0b_1,k_0c_1\not\in \mathbb{Z}$. Note that we've $\{ (k_0-1)a_1\} +\{ (k_0-1)b_1\} +\{ (k_0-1)c_1\} \leq 1$. The required inequality is true since for any $d\in \{ a,b,c\}$, we've $$\{ k_0d_1\} =\{ (k_0-1)d_1+d_1\} \leq \{ (k_0-1)d_1\} +\{ d_1\} =\{ (k_0-1)d_1\} +d_1.$$Summing three inequalities gives us $\{ k_0a_1\} +\{ k_0b_1\} +\{ k_0c_1\} \leq 1+(a_1+b_1+c_1)<2$. Also, if $k_0a_1\in \mathbb{Z}$, we get that $1=\{ (k_0-1)a_1\} +a_1$. So, $$1+\{ k_0b_1\} +\{ k_0c_1\} <2\implies \{ k_0a_1\} +\{ k_0b_1\} +\{ k_0c_1\} =\{ k_0b_1\} +\{ k_0c_1\} <1,$$contradiction with the definition of $k_0$. Hence, $k_0a_1\not\in \mathbb{Z}$. Similarly, $k_0b_1,k_0c_1\not\in \mathbb{Z}$, done. If $a_1+b_1+c_1>2$, setting $a_2=1-a_1,b_2=1-b_1,c_2=1-c_1$ and use the fact that $$\{ kd_2\} =1-\{ kd_1\}$$for all $d\in \{ a,b,c\}$. This means the $k$ that works for $a_2,b_2,c_2$ will works for $a_1,b_1,c_1$ too (also $\{ ka_2\} ,\{ kb_2\} ,\{ kc_2\} \neq 0$.) Since $a_2+b_2+c_2=3-(a_1+b_1+c_1)<1$ and $a_2,b_2,c_2\in (0,1)$, this already in the former case, done.