3. The number of triples of positive integers $(x, y, z)$ that satisfy $x y z=3^{2010}$ and $x \leqslant y \leqslant z<x+y$ is $\qquad$.


3. 336 .

Let $x=3^{a}, y=3^{b}, z=3^{c}$. Then
$$
0 \leqslant a \leqslant b \leqslant c, a+b+c=2010 \text {. }
$$

If $c \geqslant b+1$, then
$$
x+y=3^{a}+3^{b}<3^{b+1} \leqslant 3^{c}=z \text {, }
$$

which contradicts $z<x+y$. Hence, $c=b$.
Thus, $a+2 b=2010$.
Therefore, $670 \leqslant b \leqslant 1005$.
Hence, the number of such positive integer triples, which is the number of $b$, is $1005-670+1=336$.