∵椭圆C1:
x2
3+
y2
2=1的左右焦点为F1,F2,
∴F1(-1,0),F2(1,0),直线l1:x=-1,
设l2:y=t,设P(-1,t),(t∈R),M(x,y),
则y=t,且由|MP|=|MF2|,
∴(x+1)2=(x-1)2+y2,
∴曲线C2:y2=4x.
∵A(1,2),B(x1,y1),C(x2,y2)是C2上不同的点,
∴
AB=(x1−1,y1−2),
BC=(x2−x1,y2−y1),
∵AB⊥BC,
∴
AB•
BC=(x1-1)(x2-x1)+(y1-2)(y2-y1)=0,
∵x1=
1
4y12,x2=
1
4y22,
∴(y12-4)(y22-y12)+
(y1−2)(y2−y1)
16=0,
∵y1≠2,y1≠y2,
∴
(y1+2)(y1+y2)
16+1=0,
整理,得y12+(2+y2)y1+(2y2+16)=0,
关于y1的方程有不为2的解,
∴△=(2+y2)2−4(2y2+16)≥0,且y2≠-6,
∴y22−4y2−60≥0,且y2≠-6,
解得y2<-6,或y 2 ≥10.
故选:A.