3. The solution to the equation $\sqrt{x^{2}-\frac{1}{2} x+1}+\sqrt{x^{2}-\frac{2}{3} x+1}$ $=1+\frac{\sqrt{30}}{6}$ is . $\qquad$
3. $\frac{13-2 \sqrt{30}}{7}$.

The original equation can be transformed into
$$
\begin{array}{l}
\sqrt{\left(x-\frac{1}{4}\right)^{2}+\left(\frac{\sqrt{15}}{4}\right)^{2}}+\sqrt{\left(x-\frac{1}{3}\right)^{2}+\left(\frac{2 \sqrt{2}}{3}\right)^{2}} . \\
=1+\frac{\sqrt{30}}{6} .
\end{array}
$$

The left side of the equation is the sum of the distances from point $P(x, 0)$ to points $A\left(\frac{1}{4},-\frac{\sqrt{15}}{4}\right)$ and $B$ $\left(\frac{1}{3}, \frac{2 \sqrt{2}}{3}\right)$. It can be calculated that $|A B|=1+\frac{\sqrt{30}}{6}$. Therefore, point $P$ is the intersection of line $A B$ with the $x$-axis. Thus, from
$$
\frac{x-\frac{1}{4}}{\frac{\sqrt{55}}{4}}=\frac{x-\frac{1}{3}}{-\frac{2}{3} \sqrt{2}}
$$

we get $x=\frac{13-2 \sqrt{30}}{7}$.