Problem 297

(I) Since the eccentricity of the ellipse $e= \frac{1}{2}$,
We have $\frac{c}{a}=\frac{1}{2}$, thus $a=2c$,
Given $F_{2}$($c$,0),
Substitute $x=c$ into the ellipse equation, we get $y=±\frac{b^{2}}{a}$,
Thus, $|AB|=\frac{2b^{2}}{a}$,
Given the area of triangle $F_{1}AB$ is 3,
$\frac{1}{2}×2c×\frac{2b^{2}}{a}=3$,
Thus, $b^{2}=3$,
Substitute $b^{2}$ into $a^{2}-c^{2}=b^{2}$, we get $a^{2}=4$,
Thus, the equation of the ellipse is $\frac{x^{2}}{4}+ \frac{y^{2}}{3}=1$;

(II) Proof: Given P($x_{1}$,$y_{1}$), then M($-x_{1}$,$-y_{1}$), N($x_{1}$,$-y_{1}$),
The equation of the line QM is $y+y_{1}=\frac{y_{2}+y_{1}}{x_{2}+x_{1}}(x+x_{1})$,
Let $y=0$, we get $m=\frac{x_{2}y_{1}-x_{1}y_{2}}{y_{1}+y_{2}}$,
The equation of the line QN is $y+y_{1}=\frac{y_{2}+y_{1}}{x_{2}-x_{1}}(x-x_{1})$,
Let $x=0$, we get $n=\frac{x_{1}y_{2}+x_{2}y_{1}}{x_{1}-x_{2}}$,
Given R is the midpoint of PQ,
Thus, R($\frac{x_{1}+x_{2}}{2}$, $\frac{y_{1}+y_{2}}{2}$),
The slope of the line OR is $k_{OR}=\frac{y_{1}+y_{2}}{x_{1}+x_{2}}$,
Therefore, $m•n•k_{OR}=\frac{x_{2}y_{1}-x_{1}y_{2}}{y_{1}+y_{2}}•\frac{x_{1}y_{2}+x_{2}y_{1}}{x_{1}-x_{2}}•\frac{y_{1}+y_{2}}{x_{1}+x_{2}}=\frac{(x_{2}y_{1})^{2}-(x_{1}y_{2})^{2}}{x_{1}^{2}-x_{2}^{2}}=\frac{x_{2}^{2}(\frac{3}{4}a^{2}-x_{1}^{2})-x_{1}^{2}(\frac{3}{4}a^{2}-x_{2}^{2})}{x_{1}^{2}-x_{2}^{2}}=\frac{3}{4}a^{2}-3=3-3=\boxed{0}$,
Thus, $m•n•k_{OR}$ is constant.