solutions to $\frac{1}a + \frac{1}b + \frac{1}c = \frac{1}{2018}$ 
Find, with proof, all ordered triplets of positive integers $(a,b,c)$ so that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}{2018}.$

In general, if $d$ is a positive integer, then $(a,b,c) = (3d,3d,3d),(d,2d,6d),(d,6d,2d), (2d,d,6d),(2d,6d,d),(6d,d,2d),$ and $(6d,2d,d)$ are all ordered triplets of positive integers such that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}d.$ However, I'm unsure how to find all such triplets. I tried manipulating the equation as follows:
\begin{align}
&1 + \dfrac{a}b + \dfrac{a}c = \dfrac{a}{2018}\\
&\dfrac{a}b + \dfrac{a}c = \dfrac{a-2018}{2018}\\
&\dfrac{ab}{c } = \dfrac{b(a-2018)-2018a}{2018}\\
&c[b(a-2018)-2018a]-2018ab = 0\\
&c[(a-2018)(b-2018)-2018^2]-2018ab = 0\\
&c[(a-2018)(b-2018)]-2018(ab-2018(a+b)+2018^2+2018(a+b)-2018^2)-2018^2 c =0 \\
&(c-2018)(a-2018)(b-2018)-2018^2(a+b+c)+2018^3 = 0\\
&(c-2018)(b-2018)(a-2018) = 2018^2(a+b+c) - 2018^3,
\end{align}
but I don't know whether this is useful.
Source (from comments): I based this off a contest problem. I came up with it by myself. The question I based it on was the Putnam 2018 question A1.

Partial answer. $\,$ Let $d=\text{gcd}(a,b,c)$ and, thus, $a=dx, b=dy, c=dz$ - where $\text{gcd}(x,y,z)=1$. If we assume that $\text{gcd}(x,y)=\text{gcd}(y,z)=\text{gcd}(z,x)=1$ then $\text{gcd}(xyz, xy+yz+zx)=1$ $$\frac1a+\frac1b+\frac1c=\frac1{2018}\iff \frac{abc}{ab+bc+ca}=\frac{dxyz}{xy+yz+zx}=2018$$ This implies that $xy+yz+zx\mid d\iff d=(xy+yz+zx)\cdot k, k\in\mathbb Z$. Hence, the equation becomes $$k\cdot xyz=2018$$
Once you have the integer solutions to this equation - and this should not take long, since $2018=2\cdot 1009$ -, use the values to obtain back the solutions $$(a,b,c)\equiv (kx\cdot (xy+yz+zx),ky\cdot (xy+yz+zx), kz\cdot (xy+yz+zx) )$$
Observation. Due to symmetry, you "only" have to consider the case $\text{gcd}(x,y)>1$.
