25~26高三下·吉林延边一模·第7题

$8$

由直线方程$mx – y – 2m + 1 = 0$可化为$m(x – 2) – (y – 1) = 0$，
知直线恒过定点$P(2,1)$；

圆$C:(x – 1)^2 + (y – 2)^2 = 6$的圆心为$C(1,2)$，半径$r = \sqrt{6}$；

由于$|CP| = \sqrt{(2 – 1)^2 + (1 – 2)^2} = \sqrt{2} < r$，
故点$P$在圆内，直线与圆恒相交于两点$M$，$N$.

设弦$MN$的中点为$H$，则$CH \perp MN$，从而$\overrightarrow{MH} \perp \overrightarrow{HC}$，
$$
\begin{align} \overrightarrow{MN} \cdot \overrightarrow{MC} &= (\overrightarrow{MH} + \overrightarrow{HN}) \cdot (\overrightarrow{MH} + \overrightarrow{HC}) \ &= 2\overrightarrow{MH} \cdot (\overrightarrow{MH} + \overrightarrow{HC}) \ &= 2|\overrightarrow{MH}|^2 + 2\overrightarrow{MH} \cdot \overrightarrow{HC} \ &= 2|\overrightarrow{MH}|^2 = \frac{1}{2}|MN|^2, \end{align}
$$
过圆内定点$P$的弦中，当弦与$CP$垂直时弦长$|MN|$最短，
此时圆心到直线的距离$d = |CP| = \sqrt{2}$，
最短弦长为$|MN|_{\min} = 2\sqrt{r^2 – d^2} = 2\sqrt{6 – 2} = 4$.

故$\overrightarrow{MN} \cdot \overrightarrow{MC}$最小值为$\dfrac{1}{2} \times 4^2 = 8$.

故答案为：$8$.