3. Given $p_{1}, p_{2}, \cdots, p_{2015}$ are the first 2015 prime numbers in ascending order, let $m=p_{1} p_{2} \cdots p_{2015}$. Then the set of positive integer solutions for the indeterminate equation
$$
(2 x-y-z)(2 y-z-x)(2 z-x-y)=m
$$

is $\qquad$ .
3. $\varnothing$.

From the differences between $2 x-y-z$, $2 y-z-x$, and $2 z-x-y$ being multiples of 3, we know
$$
\begin{array}{c}
2 x-y-z \equiv 2 y-z-x \\
\equiv 2 z-x-y(\bmod 3) .
\end{array}
$$

From the given, $m$ is a multiple of 3.
According to equation (1), if the indeterminate equation in the problem holds, then
$$
2 x-y-z, 2 y-z-x, 2 z-x-y
$$

are all multiples of 3, meaning the left side of the original equation has a factor of $3^{3}$, but $m$ is not divisible by $3^{3}$, leading to a contradiction.
Therefore, the set of positive integer solutions to the original equation is $\varnothing$.