8. (2001 National Junior High School League Question) Find the values of real numbers $x$ and $y$ that minimize the function $f(x, y)=(1-y)^{2}+(x+y-$ $3)^{2}+(2 x+y-6)^{2}$.
8. Let $a=(1-y, x+y-3,-2 x-y+6), b=(1,2,1)$, then $a \cdot b=(1-y) \cdot 1+2 \cdot(x+y-3)+$ $(-2 x-y+6) \cdot 1=1$,
$$
|a| \cdot|b|=\sqrt{(1-y)^{2}+(x+y-3)^{2}+(-2 x-y+6)^{2}} \cdot \sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{f(x, y)} \cdot \sqrt{6} \text {. }
$$

And $a \cdot b \leqslant|a| \cdot|b|$,
then $1 \leqslant \sqrt{f(x \cdot y)} \cdot \sqrt{6}$,

which means $f(x \cdot y) \geqslant \frac{1}{6}$. Equality holds if and only if $a=\lambda b(\lambda>0)$.
Thus, by $\frac{1-y}{1}=\frac{x+y-3}{2}=\frac{-2 x-y+6}{1}=\lambda>0$,

we get $x=\frac{5}{2}, y=\frac{5}{6}, \lambda=\frac{1}{6}$, i.e., when $x=\frac{5}{2}, y=\frac{5}{6}$, $f(x, y)_{\min }=\frac{1}{6}$.