3. The number of positive integer solutions to the equation $a b c=2(a+b+c)$ is ( ).
(A) 3
(B) 6
(C) 12
(D) 15
3. D.

Notice that the original equation is symmetric with respect to $a$, $b$, and $c$.
Without loss of generality, assume $a \leqslant b \leqslant c$. Then $a+b \leqslant 2 c$.
Substituting into the original equation, we get $2 < a b \leqslant 6$.
When $a b=3$, we get
$$
(a, b, c)=(1,3,8) \text {; }
$$

When $a b=4$, we get
$$
(a, b, c)=(1,4,5),(2,2,4) \text {; }
$$

When $a b=5$, we get
$(a, b, c)=(1,5,4)$ (discard);
When $a b=6$, $c$ is not an integer.
In summary, when $a \leqslant b \leqslant c$, the original equation has three sets of positive integer solutions
$$
(a, b, c)=(1,3,8),(1,4,5),(2,2,5) \text {. }
$$

Therefore, the original equation has a total of
$$
2 \times 6+3=15 \text { (sets) of positive integer solutions. }
$$