2023/011) Solve positive integers a,b,c such that abc=a+b+c

Without loss of generality assume $ a \ge b \ge c$ so $a+b+c \le 3a$

We have $abc \le 3a$ 

Or  $bc \le 3$

That is c = 1 and b =2 giving a = 3

Or c = 1 and b = 3 giving a = 2 but this violates the condition

So solution is $(3,2,1)$ or any permutation because of symmetry