3. If a two-digit number $\bar{x}$ and a three-digit number $\overline{3 y z}$ have a product of 29,400, then, $x+y+z=$ $\qquad$
3.18.

According to the problem, we have
$$
73<\frac{29400}{400}<\overline{x 5} \leqslant \frac{29400}{300}=98 .
$$

Thus, $\overline{x 5}=75$ or 85 or 95.
Since $85 \times 29400, 95 \times 29400$, it can only be $\overline{x 5}=75$.
When $\overline{x 5}=75$, $\overline{3 y z}=\frac{29400}{75}=392$.
Therefore, $x=7, y=9, z=2$.