3. The equation about $x$
$$
\frac{x-a}{b c}+\frac{x-b}{a c}+\frac{x-c}{a b}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \text {, }
$$

where $a+b+c \neq 0$, the solution of the equation is $\qquad$ .
3. $x=a+b+c$. Transform the original equation into
$$
\begin{array}{l}
\left(\frac{x-a}{b c}-\frac{1}{b}-\frac{1}{c}\right)+\left(\frac{x-b}{a c}-\frac{1}{a}-\frac{1}{c}\right)+ \\
\left(\frac{x-c}{a b}-\frac{1}{a}-\frac{1}{b}\right)=0 .
\end{array}
$$

From this, we can solve for $x=a+b+c$.