All solutions of $a+b+c=abc$ in natural numbers

Without loss of generality $a \leq b \leq c$. Then $a+b+c \leq 3c$ and hence

$$abc=a+b+c \leq 3c$$

Thus, either $c =0$, in which case $a=b=c=0$, or

$$ab \leq 3 \,.$$

This leads to only four possibilities to check: $a=0$ or $(a,b)=(1,1)$ or $(a,b)=(1,2)$ or $(a,b)=(1,3)$.

If $a=0$ then you require $b+c=0$ and hence $b=c=0$.

Note that you can assume $a\leq b \leq c$. If $a, b, c \geq 2$ then $abc \geq 4c > c + b + a$. Hence at least one of $a,b,c$ is equal to $1$.

Wlog assume $a=1$, and look for solutions to $b+c+1 = bc$. If $b,c\geq 3$ then $bc \geq 3c > b + c + 1$, hence at least one of $b,c$ is less than $3$

Wlog assume $b=2$, and look for solutions to $c+3 = 2c$, which implies $c=3$.

So the only solutions are $(0,0,0)$ and $(1,2,3)$ and their permutations.

Here's a start of a full solution: The right side grows way faster than the left side, so it's unlikely that there are very many solutions. More formally, suppose that $a, b, c \ge 2$, and that $c$ is at least as large as $a, b$. Then we have

$$abc \ge 4c > c + c + c \ge c + b + a$$

so it's necessary that one of the numbers (which we'll call $a$) is $1$. So we can reduce the problem to studying

$$b + c = bc - 1$$

which has fewer variables.