3. Let $M_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4)$ be four points on the parabola $y=a x^{2}$ $+b x+c(a \neq 0)$. When $M_{1}, M_{2}, M_{3}, M_{4}$ are concyclic, $x_{1}+x_{2}+x_{3}+x_{4}=$ $\qquad$ .
3. $-\frac{2 b}{a}$.
$M_{2}, M_{2}, M_{3}, M_{6}$ are concyclic. Let's assume the equation of this circle is $x^{2} + y^{2} + D x + E y + F = 0$. Combining this with $y = a x^{2} + b x + c$ and eliminating $y$, we get $a^{2} x^{4} + 2 a b x^{3} + (2 a c + E a + 1) x^{2} + (2 b c + D + E b) x + c^{2} + E c + F = 0$. It is easy to see that the four roots of this equation are $x_{1}, x_{2}, x_{3}, x_{4}$. By Vieta's formulas, $x_{1} + x_{2} + x_{3} + x_{4} = -\frac{2 b}{a}$.