125 Let $f(x)=1+2 \cos x+3 \sin x$, if for any $x$ the identity $a f(x)+b f(x-c)$ $=1$ holds, then $(a, b, c)=$ $\qquad$ .
$125\left(\frac{1}{2}, \frac{1}{2},(2 n+1) \pi\right)(n \in \mathbf{Z})$
Substituting $x=0, \frac{\pi}{2},-\frac{\pi}{2}$ into the identity, we solve from the resulting equations that $a+b=1, \sin c=0$, $\cos c=-\frac{a}{b}$. Thus, $c=(2 n+1) \pi$ (where $n$ is an integer), $a= \pm b$, and therefore $a=b=\frac{1}{2}$. These values satisfy the original identity.