$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\left((\mathrm{a}+\mathrm{b}+\mathrm{c})^{2}-\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)\right) / 2=183$, so $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are roots of the cubic $\mathrm{x}^{3}-24 \mathrm{x}^{2} +183 x-440=0$. But it easily factorises as $(x-5)(x-8)(x-11)=0$, so the only solutions are permutations of $(5,8,11)$.