Need to define an integral expression to find  the area of intersection of a Plane and a Cone. 
How can the integral expression be defined  to find Area (S) on $x+y+z=1$ and  bordered with intersection of the cone ($x^2+y^2-z^2=0$) and the  plane ($x+y+z=1$) ?
 A: Using $x$ and $y$ as parameters on the plane, the parametric representation is ${\bf R} = [x,y,1-x-y]$ and $\frac{\partial {\bf R}}{\partial x} \times \frac{\partial {\bf R}}{\partial z} = [1,1,1]$ has length $\sqrt{3}$, so the area element is $\sqrt{3} \ dx \ dy$. 
As I wrote in my comment, the curve of intersection with the double cone has $(x-1)(y-1)=1/2$.  Since e.g. the point $(0,0,1)$ is inside the cone and has $(x-1)(y-1)=1 > 1/2$, the region inside the cone is $(x-1)(y-1) \ge 1/2$.  This has two components, one where $x > 1$ and $y  \ge 1 + \frac{1}{2(x-1)} = \frac{2x-1}{2x-2}$ (and $z < 0$) and one where $x < 1$ and $y \le \frac{2x-1}{2x-2}$ (and $z > 0$).
So if you still want to set up the area as an integral, it is
$$ \int_1^\infty \ dx \int_{\frac{2x-1}{2x-2}}^\infty \ dy \ \sqrt{3}
+ \int_{-\infty}^1 \ dx \int_{-\infty}^{\frac{2x-1}{2x-2}} \ dy \ \sqrt{3}$$
Both integrals, of course, are $+\infty$.
Here's a picture of part of the cone and plane.

A: The angle between the $z$ axis and the cone is the same as the angle between the $z$ axis and a plane normal to $(1,0,1)$. Without explicitly calculating this angle, you can see that it must be less than the angle between the $z$ axis and a plane normal to $(1,1,1)$. Thus, as Robert commented, the intersection is not an ellipse, as your image suggests, but a hyperbola, which has infinite area.
