Any solutions is either in the following set, or is some permutation of an element in this set:

$\displaystyle \{(m,2m,3m) \mid m\in \mathbb{Z}, m\ge 0\}\cup \{(-m,m,mn) \mid m,n\in \mathbb{Z}, m>0, n\ge 0\}$

It is a little tricky to prove that these are the only solutions, though.

Here is a sketch of a proof that if $\displaystyle 0<a\le b\le c$, where $\displaystyle \text{LCM}(a,b,c) = a+b+c$, then $\displaystyle (a,b,c) = (a,2a,3a)$:

Suppose $\displaystyle 0 < a \le b \le c$ and $\displaystyle \text{LCM}(a,b,c) = a+b+c$. Then suppose $\displaystyle m$ divides any two of the numbers. Without loss of generality, assume $\displaystyle m | a$ and $\displaystyle m | b$. Then since $\displaystyle a | \text{LCM}(a,b,c)$, we know $\displaystyle m | \text{LCM}(a,b,c)$. But then since $\displaystyle m$ divides both $\displaystyle a,b$, we have some integer $\displaystyle j$ such that $\displaystyle a+b = mj$. Then $\displaystyle m | \text{LCM}(a,b,c) = a+b+c = mj+c$ implies $\displaystyle m | c$. Hence, if a number divides two of the numbers, it must also divide the third.

Hence, any solution where $\displaystyle a,b,c$ are all positive must be a multiple of some solution $\displaystyle (a',b',c')$ where all three are pairwise coprime. But then $\displaystyle \text{LCM}(a',b',c') = a'+b'+c' = a'b'c'$. Since $\displaystyle a\le b\le c$, it must be that $\displaystyle a'\le b'\le c'$. Since these numbers are pairwise coprime, we can only have equality if $\displaystyle a'=b'=1$ or $\displaystyle a'=b'=c'=1$. Since $\displaystyle \text{LCM}(1,1,c) = c \neq 1+1+c = c+2$, it cannot be that $\displaystyle a'=b'=1$. Hence, $\displaystyle a' < b' < c'$, so $\displaystyle a'+b'+c' < 3c'$. Then $\displaystyle a'b'c' = a'+b'+c' < 3c' \Rightarrow a'b' < 3$. Since $\displaystyle a'b' \neq 1$, it must be $\displaystyle a'b' = 2$, so $\displaystyle a'=1,b'=2$. Then $\displaystyle 2c' = 1+2+c'$ implies $\displaystyle c'=3$. Hence, the only solution with all three numbers positive and coprime is $\displaystyle (1,2,3)$, so the set of all solutions with all three numbers positive is $\displaystyle \{(m,2m,3m)\mid m\in \mathbb{Z}, m>0\}$. Since $\displaystyle (0,0,0)$ is a solution, we can include $\displaystyle m=0$.