One. (Full marks 20 points) In the Cartesian coordinate system, there are three points $A(0,1), B(1,3), C(2,6)$. It is known that the points on the line $y=a x$ $+b$ with x-coordinates $0, 1, 2$ are $D, E, F$ respectively. Try to find the values of $a, b$ that minimize $A D^{2}+B E^{2}+C F^{2}$.
$F(2,2 a+b)$. From the graph, we can see
$$
\begin{array}{l} 
A D^{2}+B E^{2}+C F^{2} \\
=(b-1)^{2}+(a+b-3)^{2}+(2 a+b-6)^{2} \\
= 5 a^{2}+6 a b+3 b^{2}-30 a-20 b+46 \\
= 5 a^{2}+(6 b-30) a+3 b^{2}-20 b+46 \\
= 5\left(a+\frac{3}{5} b-3\right)^{2}-5\left(\frac{3}{5} b-3\right)^{2}+3 b^{2}-20 b \\
+46 \\
= 5\left(a+\frac{3}{5} b-3\right)^{2}+\frac{6}{5} b^{2}-2 b+1 \\
= 5\left(a+\frac{3}{5} b-3\right)^{2}+\frac{6}{5}\left(b-\frac{5}{6}\right)^{2}+\frac{1}{6} .
\end{array}
$$

When $\left\{\begin{array}{l}a+\frac{3}{5} b-3=0, \\ b-\frac{5}{6}=0\end{array}\right.$, the above expression achieves its minimum value.
At this point, $a=\frac{5}{2}, b=\frac{5}{6}$. The minimum value is $\frac{1}{6}$.