3. Given real numbers $a, b, c$ satisfy $\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}$ $=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|$. If the quadratic function $y=a x^{2}+b x$ $+c$ intersects the $x$-axis at a fixed point, then the coordinates of this fixed point are $\qquad$.
3. $(1,0)$.

From the given, we have
$$
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2},
$$

which implies $\square$
$$
\begin{array}{l}
\frac{2}{a b}+\frac{2}{b c}+\frac{2}{c a}=0 . \\
\therefore a+b+c=0 . \\
\because y=a x^{2}+b x+c \text { intersects the } x \text { axis, } \\
\therefore a x^{2}+b x+c=0 .
\end{array}
$$

From (1) and (2), we know that equation (2) must have a root of 1. Therefore, the graph of $y=a x^{2}+b x+c$ passes through the fixed point $(1,0)$.