The area over which you now need to integrate is the interior of the triangle with vertices A= (0,0,1), B= (2,0,9), and C= (2,2,11). The vector from A to B is <2, 0, 8> and from A to C is <2, 2, 10>. A normal to that plane is given by the cross product: <-16, -2, 4> or, dividing by 2 to simplify, <8, -1, 2>. The equation of the plane is 8x- y+ 2(z-1)= 0 or 8x- y+ 2z= 2 which we can finally write as z= 1- 4x+ (1/2)y. Writing that as a vector equation, using x and y as parameters, \vec{r}(x,y)= x\vec{i}+ y\vec{j}+ (1- 4x+ y/2)\vec{i}. Then \vec{r}_x= \vec{i}- 4\vec{k} and \vec{r}_y= \vec{j}+ 1/2\vec{k}. The "fundamental vector product for the surface is the cross product of those, 2\vec{i}- (1/2)\vec{j}+ \vec{k}. You need to integrate the dot product of that with curl F with respect to x and y, over the area in the xy-plane with vertices (0,0), (2,0), and (2,2), the projection of the points given to the xy-plane.
