Apply Stokes theorem to prove that $\int_{c} ydx+zdy+xdz =-2\sqrt{2}\pi a^2$
Where C is the curve given by $x^2+y^2+z^2-2ax-2ay=0, x+y=2a$ ; and begins at the point (2a,0,0) and it goes first below the z plane.
So, I got F = x$\hat{i}$ + y$\hat{j}$ +z$\hat{k}$ and then I used -$\int\int_{c} (∇\times F).n ds$ and my answer is coming($2\sqrt{2}\pi a^2$) which is negative of what is to prove and i think its correct because they have said the curve goes below z plane first, that means it's oriented in negative direction.
Too long for a comment:
The equation $$ x^2+y^2+z^2-2ax-2ay=0 $$ can be written as $$ (x-a)^2+(y-a)^2+z^2=2a^2 $$ and thus describes a sphere around $(a,a,0)$ with radius $r=\sqrt{2}\,|a|\,.$ Your curve $C$ is the intersection of that sphere with the plane $$ x+y=2a $$ which is parallel to the $z$-axis, goes through the center of the sphere and has unit normal vector $$ \mathbf{n}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\end{pmatrix}\,. $$ If you use the right hand rule you see that the orientation of this is compatible with saying that the curve starts at $(2a,0,0)$ and goes below the $x$-$y$-plane first.
The vector field $F=(y,z,x)$ you are supposed to integrate over $C$ has curl $(-1,-1,-1)\,.$ The dot product of that curl with $\mathbf{n}$ is the negative constant $-\sqrt{2}$.
Stokes tells you that we can integrate that constant over the disk $B$ with radius $r=\sqrt{2}\,|a|\,,$ center $(a,a,0)$ and lying in the plane $x+y=2a\,:$ $$ \int_C F\cdot\,ds=\int_B {\rm curl} F\cdot\mathbf{n}\,dS=-\sqrt{2}\,\pi\,r^2=-2\sqrt{2}\,\pi\, a^2\,. $$