Example 6 Find the values of real numbers $x$ and $y$ such that $(y-1)^{2}+(x+y-3)^{2}+(2 x+y-6)^{2}$ reaches the minimum value.
Let $\vec{a}=(y-1, x+y-3,2 x+y-6), \vec{b}=(-1,2,-1)$.
Since $|\vec{a}|^{2}=(y-1)^{2}+(x+y-3)^{2}+(2 x+y-6)^{2}$
$$
\begin{array}{l}
\geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(-y+1+2 x+2 y-6-2 x-y+6)^{2}}{(-1)^{2}+2^{2}+(-1)^{2}} \\
=\frac{1}{6},
\end{array}
$$

where the equality holds if and only if $\frac{y-1}{-1}=\frac{x+y-3}{2}=\frac{2 x+y-6}{-1}$, i.e., $x=\frac{5}{2}, y=\frac{5}{6}$. Therefore, $(y-1)^{2}+(x+y-3)^{2}+(2 x+y-6)^{2}$ attains its minimum value $\frac{1}{6}$.