Finding all integer solutions of $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12$

If we have integers $$x,y,z$$ such that $$x,y,z \ge 3,$$ find all solutions to $$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12.$$

I was thinking of first expanding out this, and then simplifying from there, but the equation got very messy. Is there a better method?

WLOG let $$x\leq y$$. Then, we have that $$x = 3$$ because we are given that $$x\geq 3$$, and if $$x\geq 4$$, $$\frac{1}{x}+\frac{1}{y}\leq \frac{1}{2}$$.

Then, we must solve $$\frac{1}{y}-\frac{1}{z} = \frac{1}{6}$$. Note that $$y = 3,4,5$$ because they are the only values such that $$y\geq 3$$ and $$\frac{1}{y}>\frac{1}{6}$$. Then, we can go into casework:

For $$y = 3$$, we get $$\frac{1}{z} = \frac{1}{6}$$ and thus $$z = 6$$.

For $$y = 4$$, we get $$\frac{1}{z} = \frac{1}{12}$$ and thus $$z = 12$$.

Finally, for $$y = 5$$ we get $$\frac{1}{z} =\frac{1}{30}$$ and thus $$z = 30$$.

We must permute $$x$$ and $$y$$ when they are different to get all of the solutions because of the earlier WLOG.

Thus, the only solutions are $$\boxed{(3,3,6),\ (3,4,12),\ (3,5,30),\ (4,3,12),\text{ and }(5,3,30).}$$

As an alternative to Joshua Wang's excellent answer:

By symmetry we may assume, without loss of generality, that $$x\leq y$$. Clearing denominators yields $$2xz+2yz-2xy=xyz,\tag{1}$$ and a bit of rearranging then shows that $$xy(z+2)=2(x+y)z.$$ Of course $$z+2>z$$ and so $$xy<2(x+y)$$, or equivalently $$(x-2)(y-2)<4.$$ Because $$x\leq y$$ we see that $$x<4$$, so $$x=3$$ and then $$y<6$$. Plugging into $$(1)$$ and rearranging shows that $$yz+6y-6z=0,$$ or equivalently $$z=\frac{6y}{6-y}$$, yielding the following three solutions for $$(x,y,z)$$: $$(3,3,6),\qquad(3,4,12),\qquad(3,5,30).$$