Evaluate $\ \int_T \vec F \cdot d\vec r $ where $\vec F (x, y, z) = (2xy + 4xz)\vec i + (x^2 + 6yz)\vec j + (2x^2 + 3y^2) \vec k$ I'm a bit confused as to what I've to do for this exam question.  
$$\vec F (x, y, z) = (2xy + 4xz)\vec i + (x^2 + 6yz)\vec j + (2x^2 + 3y^2) \vec k , \qquad x, y, z ∈ \mathbb{R}. $$
Let $T$ denote the triangular path with vertices
$(1, 1, 1)$, $(2, 1, 1)$, and $(3, 2, 2)$,
traversed from $(1, 1, 1)$ to $(2, 1, 1)$ to $(3, 2, 2)$ to $(1, 1, 1)$. Evaluate
$$\ \int_T \vec F \cdot d\vec r. $$
Justify your answer.
I was thinking of parameterising with a set of lines but I'm not too sure how I would go about this? 
Thanks for any help you can give 

Note that we can find a potential function $\phi$ for which we have $\vec F(x,y,z)=\nabla \phi(x,y,z)$.  To find $\phi$, we need only integrate the components of $\vec F$.
To proceed, we begin with the $\hat x$-component to find
$$\begin{align}
\phi(x,y,z)&=\int F_x(x,y,z)\,dx\\\\
&=\int (2xy+4xz)\,dx\\\\
&=x^2y+2x^2z+C_1(y,z) \tag 1
\end{align}$$
where $C_1(y,z)$ is an integration constant with respect to $x$, but can depend on both $y$ and $z$.
Next, we use $\phi(x,y,z)$ as given in $(1)$ to find $C_1(y,z)$.  Noting that $\frac{\partial \phi(x,y,z)}{\partial y}=F_y(x,y,z)$ yields
$$x^2+\frac{\partial C_1(y,z)}{\partial y}=x^2+6yz$$
from which we find $C_1(y,z)=3y^2z+C_2(z)$, where $C_2(z)$ is a second integration constant.  At this point, we can write the potential function as
$$\phi(x,y,z)=x^2y+2x^2z+3y^2z+C_2(z) \tag 2$$
Finally, taking the partial of $\phi(x,y,z)$, as given by $(2)$, with respect to $z$, and setting it equal to $F_z(x,y,z)$ yields
$$2x^2+3y^2+C_2'(z)=2x^2+3y^2$$
from which we find that $C_2(z)$ is a constant.  
Therefore, we can write the potential function as
$$\bbox[5px,border:2px solid #C0A000]{\phi(x,y,z)=x^2y+2x^2z+3y^2z+C}$$
for any constant $C$.

Since we have determined that $\vec F(x,y,z)=\nabla \phi(x,y,z)$, $\vec F$ is conservative, and its path integral depends only on the end points of the path.  We can write
$$\begin{align}
\int_{\vec r_1}^{\vec r_2}\vec F(x,y,z)\cdot \vec d\vec \ell&=\int_{\vec r_1}^{\vec r_2} \nabla \phi(x,y,z)\cdot d\vec \ell\\\\
&=\int_{\vec r_1}^{\vec r_2} d\phi\\\\
&=\phi(1,1,1)-\phi(1,1,1)\\\\
&=0\end{align}$$       
