Fermat Triangle

First we notice that $\triangle ABC$ has no angle $\ge 120{}^\circ$ , otherwise, the Fermat Point would be the vertex of the obtuse angle. Consequently, the Fermat Point of $\triangle ABC$ is the 1st Isogonic Center of the triangle, i.e. the angles subtended by the sides of the triangle at $F$ are all equal, thus,

$\angle AFB=\angle BFC=\angle CFA=120{}^\circ$

Using cosine rule on $\triangle AFB$ , $\triangle BFC$ and $\triangle CFA$ , we have

$A{{B}^{2}}=A{{F}^{2}}+B{{F}^{2}}-2AF\cdot BF\cdot \cos (\angle AFB)={{5}^{2}}+{{3}^{2}}-2\times 5\times 3\times \left( -\dfrac{1}{2} \right)=49\Rightarrow AB=\sqrt{49}$

$B{{C}^{2}}=B{{F}^{2}}+C{{F}^{2}}-2BF\cdot CF\cdot \cos (\angle BFC)={{3}^{2}}+{{4}^{2}}-2\times 3\times 4\times \left( -\dfrac{1}{2} \right)=37\Rightarrow BC=\sqrt{37}$

$C{{A}^{2}}=C{{F}^{2}}+A{{F}^{2}}-2CF\cdot AF\cdot \cos (\angle CFA)={{4}^{2}}+{{5}^{2}}-2\times 4\times 5\times \left( -\dfrac{1}{2} \right)=61\Rightarrow CA=\sqrt{61}$

Thus, the perimeter of $\triangle ABC$ is $\sqrt{49}+\sqrt{37}+\sqrt{61}$ .

For the answer, $a=49$ , $b=37$ , $c=61$ , thus, $a+b+c=\boxed{147}$ .