We consider the cube $L=[0,r_0]^3$ where $r_0 > 0$ is a fixed real number.
The lattice is generated by the level set $L_r$
We define the level function $\lambda (x,y,z) \to x+y+z =r$
For $ r \leq r_0$
$$L_r = \{x \in L : \lambda(r) = r\}$$ is an equilateral triangle with vertices $(r,0,0), (0,r,0),(0,0,r)$.
For $r_0 < r < 2r_0$ the level set $L_r$ is a hexagon with vertices $$(r_0, r-r_0,0), (0,r_0,r-r_o), (0,r - r_0, r_0),(r-r_0,r_0,0), (r-r_0, 0, r_0)\;.$$
Define $\mu_L (r)$ to represent the surface area of each level set.
Therefore $$\mu_L (r) = \begin{cases} \frac{\sqrt3}{2}r^2, & \text{if } 0\leq r \leq r_0 \\\\ \frac{\sqrt3}{2}(r^2-3(r-r_0)^2), & \text{if } r_0 < r \leq 2r_0 \\\\ \frac{\sqrt3}{2}(3r_0-r^2), & \text{if }2r_0 \leq r \leq 3r_0 \\ \end{cases} $$
Can someone kindly explain what is going on here? What is a level set? Where do the second and third premises of $\mu_l (r)$ come from? What is a level function?
Let’s look at a $2$-dimensional version, and let me replace $r_0$ by $1$: consider the square $L=[0,1]^2$. Every point of $L$ lies on a line $x+y=r$ for some $r$ with $0\le r\le 2$. The origin is the only of $L$ on the line $x+y=0$, and the point $\langle 1,1\rangle$ is the only point of $L$ on the line $x+y=2$. The line $x+y=1$ contains the diagonal of the square $L$ running from upper left to lower right. I strongly recomment that you make a sketch.
For $p=\langle x,y\rangle\in L$ let $\lambda(p)=x+y$; call $\lambda(p)$ the level of $p$. If $p=\langle a,b\rangle\in L$, then $p$ lies on the line $x+y=\lambda(p)$. For $0\le r\le 2$ let $$L_r=\{p\in L:\lambda(p)=r\}=\{\langle x,y\rangle\in L:x+y=r\}\;;$$ $L_r$ is just the set of points of $L$ lying on the line $x+y=r$ and is a line segment of slope $-1$. We call these line segments the level sets of $L$.
When $0\le r\le 1$, the level set $L_r$ runs from the left edge of $L$ at the point $\langle 0,r\rangle$ to the bottom edge of $L$ at the point $\langle r,0\rangle$. When $1\le r\le 2$, however, $L_r$ runs from the top edge of $L$ at the point $\langle r-1,1\rangle$ to the righthand edge of $L$ at the point $\langle 1,r-1\rangle$. (Again, you should be drawing sketches.)
For $0\le r\le 2$ Define $\mu_L(r)$ to be the length of the level set $L_r$. Then
$$\mu_L(r)=\begin{cases} \sqrt2 r,&\text{if }0\le r\le 1\\ \sqrt2(2-r),&\text{if }1\le r\le 2 \end{cases}$$
by the Pythagorean theorem.
Now try to modify this $2$-dimensional version by letting $L=[0,r_0]^2$ for some $r_0>0$; the levels will now run from $r=0$ to $r=2r_0$. Once you’ve made that work, look again at the $3$-dimensional version in the question. $L$ is now a cube. The level set $L_r$ is the intersection with that cube of the plane $x+y+z=r$, a plane perpendicular to the diagonal line $x=y=z$. When $0\le r\le r_0$ or $2r_0\le r\le 3r_0$, the plane intersects the cube in an equilateral triangle. (It may help to find or make a cubical block and sketch on its surface the its intersections with a couple of these planes.) When $r_0<r<2r_0$, however, you’ll find that the plane actually intersects six edges of the cube: the six edges that have neither the origin nor the point $\langle r_0,r_0,r_0\rangle$ as an endpoint. For $r$ in this range the level set $L_r$ is a hexagon. (It’s a regular hexagon only when $r=\frac32r_0$, however.) The function $\mu_L(r)$ gives the area of the triangle or hexagon forming the level set at level $r$, just as my simpler function $\mu_L(r)$ gives the length of the level set $L_r$ in the $2$-dimensional example.