$\because S_{\triangle PQR}=3S_{\triangle PFO}\quad PO=OR,$
$容易得到三角形以PO和PR为底的两个三角形高之比为3:2\Rightarrow \overrightarrow{PF} =2\overrightarrow{FQ}$
$设P(x_1,y_1),Q(x_2,y_2),F(-1,0)\quad 根据定比分点公式有:$
$\begin{cases} -1=\cfrac{x_1+\lambda x_2}{1+\lambda}=\cfrac{x_1+2x_2}{1+2} \\ 0=\cfrac{y_1+\lambda y_2}{1+\lambda}=\cfrac{y_1+2y_2}{1+2}\end{cases}$
$代入曲线C有:\begin{cases} \cfrac{x_1^2}{4} +\cfrac{y_1^2}{3}=1 \quad ① \\\cfrac{x_2^2}{4} +\cfrac{y_2^2}{3}=1\quad ② \end{cases}$
$①-\lambda ^2② \Rightarrow \cfrac{1}{4}(x_1^2-\lambda ^2x_2^2)+\cfrac{1}{3}(y_1^2-\lambda ^2y_2^2)=1-\lambda ^2$
$\lambda=2,\Rightarrow -\cfrac{1}{4}(x_1-x_2)=1-2\Rightarrow x_1-2x_2=4$
$x_1+2x_2=-3,解得x_1=\cfrac{1}{2},代回①解得\cfrac{1}{3}y_1^2=1-\cfrac{1}{16},y_1=\cfrac{3}{4}\sqrt{5}$
$故直线l的斜率k=\cfrac{\cfrac{3}{4}\sqrt{5}-0}{\cfrac{1}{2}-(-1)}=\cfrac{\sqrt{5}}{2}$
$所以直线l的方程为:y=\cfrac{\sqrt{5}}{2}(x+1)$
$求\tan \angle PQR的最小值，即求k_{PQ}+k_{QR}的最小值$
$设P(x_1,y_1)Q(x_2,y_2),R(-x_1,-y_1),即\cfrac{y_2-y_1}{x_2-x_1}+\cfrac{y_2+y_1}{x_2+x_1}的最小值$
$\cfrac{y_2-y_1}{x_2-x_1}+\cfrac{y_2+y_1}{x_2+x_1}=\cfrac{(y_2-y_1)(x_2+x_1)+(y_2+y_1)(x_2-x_1)}{x_2^2-x_1^2}=\cfrac{2x_2y_2-2x_1y_1}{(x_1+x_2)(x_1-x_2)}$