A triangle $A B C$ lying in the first quadrant has two vertices as $\mathrm{A}(1,2)$ and $\mathrm{B}(3,1)$. If $\angle \mathrm{BAC}=90^{\circ}$, and $\operatorname{ar}(\triangle A B C)=5 \sqrt{5}$ sq. units, then the abscissa of the vertex $C$ is :
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