3. Let the real number $a$ be such that the quadratic equation $5 x^{2}-5 a x+66 a-1715=0$ has two integer roots. Then all such $a$ are $\qquad$

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3.870.

Let the two integer roots be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$. By the relationship between roots and coefficients, we have $a=x_{1}+x_{2}$, thus, $a$ is an integer. From the original equation, we get
$$
\begin{array}{l}
a=\frac{5 x^{2}-1715}{5 x-66}=x+13+\frac{x-857}{5 x-66} \in \mathbf{Z} \\
\Leftrightarrow \frac{x-857}{5 x-66} \in \mathbf{Z} \\
\Leftrightarrow \frac{5(x-857)}{5 x-66} \in \mathbf{Z} \text { (since } 5 \text { and } 5 x-66 \text { are coprime) } \\
\Leftrightarrow \frac{-5 \times 857+66}{5 x-66} \in \mathbf{Z} \Leftrightarrow \frac{4219}{5 x-66} \in \mathbf{Z} \\
\Leftrightarrow 5 x-66= \pm 1 \text { or } \pm 4219 \text { (since } 4219 \text { is a prime number) } \\
\Leftrightarrow x=13 \text { or } 857 .
\end{array}
$$

Therefore, $a=13+857$ or $13+13$ or $857+857$, which means $a=870$ or 26 or 1714.
Since the equation has integer roots, $5 \mid a$, which contradicts $a=26,1714$. Hence, $a=870$.