3. The product of all integers $m$ that make $m^{2}+m+7$ a perfect square is ( ).
(A) 84
(B) 86
(C) 88
(D) 90
3. A.

Let $m^{2}+m+7=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $m^{2}+m+7-k^{2}=0$.
Solving for $m$, we get $m=\frac{-1 \pm \sqrt{4 k^{2}-27}}{2}$.
Since $m$ is an integer, we should have $4 k^{2}-27=n^{2}\left(n \in \mathbf{N}_{+}\right)$, which means $(2 k+n)(2 k-n)=27$.

Solving, we get $\left\{\begin{array}{l}n=13, \\ k=7\end{array}\right.$ or $\left\{\begin{array}{l}n=3, \\ k=3 .\end{array}\right.$
Thus, $m_{1}=-7, m_{2}=6, m_{3}=-2, m_{4}=1$.
Therefore, $m_{1} m_{2} m_{3} m_{4}=84$.