10. Given that $m, n$ are constants, and the parabola $C: y=\left(t^{2}+t+\right.$ 1) $x^{2}-2(m+t)^{2} x+\left(t^{2}+3 m t+n\right)$, passes through the point $P(1,0)$ for any $t \in \mathbf{R}$.
( I ) Find the values of $m, n$;
(II) For what value of $t$ is the length of the chord intercepted by the parabola $C$ on the $x$-axis maximized? What is the maximum value?
10. ( I ) Since the parabola $C$ passes through the point $P(1,0)$, we know $x=1, y=0$, substituting into the equation of $C$ gives:
$(1-m) t+1-2 m^{2}+m=0$.
Given that the equation $D$ is independent of $t$, then $\left\{\begin{array}{l}1-m=0 \\ 1-2 m^{2}+n=0\end{array} \Rightarrow\left\{\begin{array}{l}m=1 \\ n=1\end{array}\right.\right.$.
(II) The equation of the curve $C$ is $y=\left(t^{2}+t+1\right) x^{2}-2(1+t)^{2}+\left(t^{2}+3 t+1\right)$. Setting $y=0$, we get $\left(t^{2}+t+1\right) x^{2}-$ $2(1+t)^{2} x+\left(t^{2}+3 t+1\right)=0$.
The length of the chord intercepted by the parabola $C$ on the $x$-axis is
$l=\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$, where $x_{1}, x_{2}$ are the roots of equation (2).
By Vieta's formulas, we get $l=\frac{2}{\left|t+\frac{1}{t}+1\right|}$.
Discussion: When $t>0$, $t+\frac{1}{t} \geqslant 2, t+\frac{1}{t}+1 \geqslant 3 . \therefore l \leqslant \frac{2}{3}$;
When $t=0$, substituting into equation (2) gives $x_{1}=x_{2}=1, \therefore l=0$;
When $t<0$, $-t+\frac{1}{-t} \geqslant 2, t+\frac{1}{t} \leqslant-2, t+\frac{1}{t}+1 \leqslant-1$.
Thus $\left|t+\frac{1}{t}+1\right| \geqslant 1, \therefore l \leqslant 2$.
Therefore, $l$ has a maximum value of 2 if and only if $t=-1$.