First assume $a \ge 2$, $b \ge 2$, $c \ge 2$. W.l.o.g., let $c$ be the greatest among the three numbers. Then $a^{bc} + b^{ca} + c^{ab} \ge a^4 + b^4 + c^4 > b^4 + c^4 \ge 2b^2c^2 = 2b \cdot c \cdot bc > 3 \cdot a \cdot bc$. Thus there are no solutions in this case.

It remains to study triples that contain $1$. W.l.o.g., let $a=1$. The equation reduces to $1+b^c+c^b=3bc$. Assume $b \ge 3$, $c \ge 3$. W.l.o.g., $c \ge b$, leading to $1+b^c+c^b \ge 1+b^3+c^3 > c^3 \ge 3 \cdot b \cdot c$. Thus there are no solutions in this case either. Now assume $b \ge 2$, $c \ge 2$ and one of the numbers is $2$. W.l.o.g. let $b=2$. The equation reduces to $1+2^c+c^2=6c$ which can be interpreted as a quadratic equation w.r.t. $c$ that leads to $c = 3 \pm \sqrt{9-(2^c+1)}$. Hence $8-2^c$ is a perfect square. The only candidates for this are $4$ and $0$ that give $c=2$ and $c=3$, respectively, but $c=2$ leads to contradiction (the above formula would give $c=1$ or $c=5$). The case $c=3$ gives the solution $(1,2,3)$ of the original equation. By symmetry, also $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$, $(3,2,1)$ are solutions. If one of the numbers $b$ and $c$ is $1$ then, w.l.o.g., $b=1$. The equation reduces to $1+1+c=3c$, whence $c=1$. This gives the trivial solution $(1,1,1)$.