166. The sum of the reciprocals of three positive integers is equal to one. What are these numbers?
166. First, we establish that at least one of the three numbers $x, y$, and $z$, such that

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$

must be less than 4: if all these numbers were $\geqslant 4$, then the sum $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ would be $\leqslant \frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$. Thus, if we assume that $x \leqslant y \leqslant z$, then for $x$ only two values are possible: $x=2$ and $x=3$ (since $x>1$). Let's consider these two possibilities separately.

1) $x=2$; then $\frac{1}{y}+\frac{1}{z}=1-\frac{1}{x}=\frac{1}{2}$. Bringing the fractions to a common denominator and discarding the denominator, we get:

$$
y z-2 y-2 z=0, \quad y z-2 y-2 z+4=4
$$

or

$$
(y-2)(z-2)=4
$$

Since $y$ and $z$ are greater than 1, $y-2$ and $z-2$ cannot be negative, and the following cases are possible:

A) $y-2=2, z-2=2 ; y=4, z=4$.

B) $y-2=1, z-2=4 ; y=3, z=6$.
2) $x=3$; then $\frac{1}{y}+\frac{1}{z}=1-\frac{1}{x}=\frac{2}{3}$, or

$$
2 y z-3 y-3 z=0, \quad 4 y z-6 y-6 z+9=9, \quad(2 y-3)(2 z-3)=9
$$

Since $y \geqslant x=3,2 y-3 \geqslant 3,2 z-3 \geqslant 3$, only one case is possible:

$$
2 y-3=3,2 z-3=3 ; \quad y=3, z=3
$$

Thus, all solutions to the problem are given by the following equalities:

$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1 ; \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 ; \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1
$$