$$\oint ({\mathcal{\hat{r}}} \cdot \vec r') \mathrm{ d\vec l'} = -{\mathcal{\hat{r}}} \times \int \mathrm{d\vec a'}$$
Here the integration in the LHS is around a certain loop and the $d\vec a'$ represents any surface enclosed by the loop.$\vec r' $ is the vector from origin to a point on the loop and $\mathcal{\hat{r}}$ is constant. I am unable to prove the integral.
OK, I think you know the stokes theorem which states
$$\oint\limits_{\partial S} {{\bf{F}} \cdot d{\bf{x}}} = \int\!\!\!\int\limits_S {\nabla \times {\bf{F}} \cdot d{\bf{a}}}\tag{1}$$
Now, consider the following
$$\eqalign{ & {\bf{F}} = f({\bf{x}})\,{{\bf{e}}_i},\,\,\,\,\,\,\,i = 1,2,3 \cr & \nabla \times {\bf{F}} = \nabla \times \left( {f({\bf{x}})\,{{\bf{e}}_i}} \right) = \nabla f({\bf{x}}) \times \,{{\bf{e}}_i} + f({\bf{x}})\nabla \times {{\bf{e}}_i} = \nabla f({\bf{x}}) \times {{\bf{e}}_i} \cr & \left( {\nabla \times {\bf{F}}} \right) \cdot d{\bf{a}} = \left( {\nabla f({\bf{x}}) \times {{\bf{e}}_i}} \right) \cdot d{\bf{a}} = -{{\bf{e}}_i} \cdot \left( {\nabla f({\bf{x}}) \times d{\bf{a}}} \right) \cr}\tag{2}$$
where ${{\bf{e}}_i}$'s are the orthonormal Cartesian basis. According to $(1)$ and $(2)$ we have
$${{\bf{e}}_i} \cdot \left( {\oint\limits_{\partial S} {f({\bf{x}})d{\bf{x}}} + \int\!\!\!\int\limits_S {\nabla f({\bf{x}}) \times d{\bf{a}}} } \right) = {\bf{0}}\tag{3}$$
and hence we can concluded that
$$\oint\limits_{\partial S} {f({\bf{x}})d{\bf{x}}} = -\int\!\!\!\int\limits_S {\nabla f({\bf{x}}) \times d{\bf{a}}} \tag{4}$$
This equation is an alternative form of the stokes theorem. The last step is to choose
$$f({\bf{x}}) = {\bf{c}} \cdot {\bf{x}}\tag{5}$$
where ${\bf{c}}$ is some constant vector. Now considering $\nabla \left( {{\bf{c}} \cdot {\bf{x}}} \right) = {\bf{c}}$ and using $(4)$ we have
$$\oint\limits_{\partial S} {\left( {{\bf{c}} \cdot {\bf{x}}} \right)d{\bf{x}}} = - {\bf{c}} \times \int\!\!\!\int\limits_S {d{\bf{a}}} \tag{6}$$
which is what you wanted. :)