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Let us also consider the circumcircle of $△ADF$. Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of $△ABC$ which is $P$, Also, since $m\angle ADP=m\angle AFP=90^{\circ }$. $ADPF$ is cyclic, similarly, $BDPE$ and $CEPF$ are also cyclic. With this, we know that the circumcircles of $△ADF$, $△BDE$ and $△CEF$ all intersect at $P$, so $P$ is $X$. The question now becomes calculating the sum of the distance from each vertex to the circumcenter. We can calculate the distances with coordinate geometry. (Note that $XA=XB=XC$ because $X$ is the circumcenter.) Let $A=(5,12)$, $B=(0,0)$, $C=(14,0)$, $X=(x_0,y_0)$ Then $X$ is on the line $x=7$ and also the line with slope $-\frac{5}{12}$ that passes through $(2.5,6)$ (realize this is due to the fact that $XD$ is the perpendicular bisector of $AB$). $y_0=6-\frac{45}{24}=\frac{33}{8}$ So $X=(7,\frac{33}{8})$ and $XA+XB+XC=3XB=3\sqrt{7^2+{\left(\frac{33}{8}\right)}^2}=3\times \frac{65}{8}=\boxed{\frac{195}{8}}$ Remark: the intersection of the three circles is called a Miquel point.