Find the work done by the force on a curve The question is: The work done by the force 
$
\vec{F}=y\vec{i}+x\vec{j}+z\vec{k}
$
in moving from (-1, 2, 5) to (1, 0, 1) on C the curve of intersection of the paraboloid 
$
\ z = x^2+y^2\
$and the plane $ x+y=1 $.
From what I understand, the intersection of the paraboloid and the plane is just a point which is the origin. Then, how can the curve be formed?
Any suggestion, please? Thank you!

Crossing of paraboloid $z=x^2+y^2$ and plane $x+y=1$ leads to paraabolic curve. Let us express it explicitly: $y=1-x\to z=2x^2-2x+1$. Now, introduce parameterization $x=t$, then we have parametric equation of curve: 
$$
r(t)=
\begin{cases}
x=t\\
y=1-t\\
z=2t^2-2t+1
\end{cases}\quad \text{where }t\in[-1,1]
$$
Boundaries of $t$ come from points $(-1,2,5)$ and $(1,0,1)$ declared in task.
Let us calculate the work:
$$
A=\int_C\vec{F}d\vec{r}=\int^1_{-1}\vec{F}(\vec{r}(t))\vec{r}(t)^{'}dt
$$
Where $A$-work, $C$-curve.
$$
\vec{F}=y\vec{i}+x\vec{j}+z\vec{k}\to (1-t)\vec{i}+t\vec{j}+(2t^2-2t+1)\vec{k}\\
\vec{r}(t)^{'}=\left(t\vec{i}+(1-t)\vec{j}+(2t^2-2t+1)\vec{k}\right)^{'}=(\vec{i}-\vec{j}+(4t-2)\vec{k})\\
\vec{F}(\vec{r}(t))\vec{r}(t)^{'}=(1-t)-t+(2t^2-2t+1)(4t-2)=8t^3-12t^2+6t-1
$$ 
Then,
$$
A=\int^{1}_{-1}(8t^3-12t^2+6t-1)=(2t^4-4t^3+3t^2-t)|^1_{-1}=-10
$$
