Example 9 Solve the equation $\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}(x+y+z)$.
(1978 Roma Competition)
Let $\sqrt{x}=u, \sqrt{y-1}=v, \sqrt{z-2}=t, u+v+t=s$, then we can let $l: u+v+(t-s)=0$,
$\odot O: u^{2}+v^{2}=2 s-t^{2}-3$.
Since $l$ and $O$ have a common point, then $\frac{|t-s|}{\sqrt{2}} \leqslant \sqrt{2 s-t^{2}-3}$, which means $s^{2}-2 t(t+2) s+3 t^{2}+6 \leqslant 0$.
Since $s$ has real solutions, then $\Delta=[-2(t+2)]^{2}-4\left(3 t^{2}+6\right) \geqslant 0$, which means $(t-1)^{2} \leqslant 0$, hence $t=1$. At this point, $s=t+2=3, z=t^{2}+2=3$, substituting into the equations of $l$ and $\odot O$ gives $u+v=2, u^{2}+v^{2}=2$. Solving these equations yields $u=v=1$, so the solution to the original equation is $x=1, y=2, z=3$.