flue

[ori]

parallelogram S:
A=(-1,0,0)
B=(-1,1,-2)
C=(0,1,0)
D=(0,0,2)
and the normal vector of S creates angle bigger than 90 with the possitive x axis. what is the flue of the field F=(1,-1,-2) through S?

ok so flue is plane integral type 2:
SS F *n ds
S

so how can i know what is n?

[ori]

ok that's what i did next:
AB=(0,1,-2)
CB(-1,0,-2)
AB x CB (-2,2,1)
F=(1,-1,-2)

SS F*n ds = SS (-6) ds
S

but when i split it to 4 parts:
-6SSds
from 0 to 1 & from 0 to -2
etc
the sum of them all gives 0
why am i wrong?

(the right final answer is -6)