why i got different question with stoke's theorem vector field $F = <2y,xz,1>$, 
$C$ is path made by $3$ points $a<1,0,0>,\ b<0,2,0>,\ c<0,0,3>$,
compute  $$\oint_{C}\overrightarrow{F}\ d\overrightarrow{r}$$ 
I tried to prove standard way of solving and I got -2
but when I tried using stoke's theorem by using 
$$\int\int curl\overrightarrow{F}.(\overrightarrow{ca} x \overrightarrow{ cb}) dxdy$$
i got different answer.... and when I parameterized my equation into r(x,y)=
and do $$\int\int curl\overrightarrow{F}.(\overrightarrow{r_x} x \overrightarrow{r_y})dxdy$$
i also got different answer. please help me...
 A: The plane through the three given points has a cartesian equation of the form $6x+3y+2z=6$ and you can rewrite this in the form $z=f(x,y)$ with $f(x,y)=3-2x-\tfrac{3}{2}y$ and then you can use $(-f_x,-f_y,1)=\left(2,\tfrac{3}{2},1\right)$ as a normal vector.
With $F=\left( 2y,xz,1 \right)$, you have $\nabla \times F = \left( -x,0,z-2 \right)$; plugging in $\color{blue}{z=f(x,y)}$ and taking the dot product with the normal vector gives:
$$\left( -x,0,\color{blue}{3-2x-\tfrac{3}{2}y}-2 \right) \cdot \left(2,\tfrac{3}{2},1\right) = \tfrac{1}{2} \left(-12 x - 3 y + 2\right)
$$
The projection of the surface onto the $xy$-plane is a triangle, you can verify that integrating yields:
$$\int_0^1 \int_0^{2-2x} \tfrac{1}{2} \left(-12 x - 3 y + 2\right) \,\mbox{d}y\,\mbox{d}x = \int_0^1 \left( 9x^2-8x-1 \right) \,\mbox{d}x = -2$$

Addition after comments.
You can find a normal to the plane by calculating the cross product of $\vec{ca}$ and $\vec{cb}$ but you have to be careful with normalizing and take the projection of the surface onto the $xy$-plane into account. You should check your (course) notes to see what notation or procedure they use.
For a surface of the form $\color{blue}{z=f(x,y)}$ (same as above), you can use the unit normal and reducing the surface integral to a double integral over the projected region will require the introduction of an extra factor $\sqrt{1+f_x^2+f_y^2}$, so you get:
$$\iint_S \nabla \times F \cdot \vec n \,\mbox{d}S
= \iint_S \nabla \times F \cdot \frac{\vec{ca} \times \vec{cb}}{\left\| \vec{ca} \times \vec{cb} \right\|} \sqrt{1+f_x^2+f_y^2}\,\mbox{d}x\,\mbox{d}y$$
You can verify that
$$\frac{\vec{ca} \times \vec{cb}}{\left\| \vec{ca} \times \vec{cb} \right\|} \sqrt{1+f_x^2+f_y^2}$$simplifies to the $\left(2,\tfrac{3}{2},1\right)$ from above.
