$1) \overrightarrow{AB}=(\frac{8}{3},8)=>\overrightarrow{n}=(-8,\frac{8}{3})$$=>pt(d): -8(x+\frac{2}{3})+\frac{8}{3}(y+7)=0$
$y=3x-5$
2) pt hoành độ giao điểm : $-2x^2=3x-5=>x_1=1=>A(1,-2)$
$x_2=\frac{-5}{2}=>B(\frac{-5}{2},\frac{-25}{2})$
2*) theo đề ta có : $y_c+x_c=-6=>y_c=-6-x_c$
Vì C thuộc (P): $y_c=-2x_c^2=>-2x_c^2=-6-x_c=>x_{c1}=2$=>$C_1(2,-8)$
$y_{C2}=\frac{-3}{2}=>C_2=(\frac{-3}{2},\frac{-9}{2})$