Problem 923

We add $1$ on both sides of the given equality and we get
$$xyz + xy + yz + zx + x + y + z + 1 = 244$$
or
$$xy(z+1) + x(z+1) + y(z+1) + z + 1 = 244$$
or
$$(z+1)(xy + x + y + 1) = 244.$$
From where we obtain
$$(x+1)(y+1)(z+1) = 244.$$
Because $244 = 2 \cdot 2 \cdot 61$ we get that $(x+1, y+1, z+1)$ is one of the following triples $(2, 2, 61)$, $(2, 61, 2)$, $(61, 2, 2)$. Finally, the desired triples are $(1, 1, 60)$, $(1, 60, 1)$, $(60, 1, 1)$.