3. Let $p, q, r$ be prime numbers, and $\frac{p q r}{p+q+r}$ is an integer. Then ( ).
(A) $p, q, r$ must include one 2
(B) $p, q, r$ must include one 3
(C) $p, q, r$ must have two equal numbers
(D) $\frac{p q r}{p+q+r}$ is also a prime number
3. D.

Given that $p$, $q$, and $r$ are all prime numbers, and $p$, $q$, and $r$ are all less than $p+q+r$, we know that $\frac{p q r}{p+q+r}$ can only be $1$, $p$, $q$, or $r$.
If $\frac{p q r}{p+q+r}=1$, then
$$
\begin{array}{l}
p q r=p+q+r \Leftrightarrow p=\frac{q+r}{q r-1}>1 \\
\Rightarrow q r \leqslant q+r \Leftrightarrow(q-1)(r-1) \leqslant 1 \\
\Rightarrow q=r=2 \Rightarrow 3 p=4,
\end{array}
$$

which is a contradiction.
Therefore, $\frac{p q r}{p+q+r}$ can only be one of $p$, $q$, or $r$.
The other options all have counterexamples. For example, $2,3,5$ excludes option
$\mathrm{C} ; 2,5,7$ excludes option $\mathrm{B} ; 3,5,7$ excludes option $\mathrm{A}$.