Let the first term be $x$. We claim that the only possible values of $x$ are $x=12\sqrt{3}$ and $x=6\sqrt{15}-6\sqrt{3}$, which can be checked. We need to prove they are the only solutions. The perimeter of the triangle will be $x$, so the side length is $\frac{x}{3}$. Hence the next term is $\frac{\sqrt{3}}{4} \left( \frac{x}{3} \right)^2 = \frac{\sqrt{3}}{36} x^2$. Similarly, the third term will be $\frac{\sqrt{3}}{4} \left( \frac{1}{3}\cdot \frac{\sqrt{3}}{36} x^2 \right)^2=\frac{3\sqrt{3}}{36^3} x^4 = \frac{x}{8} \left( \frac{\sqrt{3}}{18} x \right)^3$ So the arithmetic progression is $x$, $\frac{\sqrt{3}}{36} x^2$, $\frac{x}{8} \left( \frac{\sqrt{3}}{18} x \right)^3$. Consequently, $$\frac{x}{8} \left( \frac{\sqrt{3}}{18} x \right)^3-\frac{\sqrt{3}}{36} x^2=\frac{\sqrt{3}}{36} x^2 - x$$$$\frac{x}{8} \left(\left( \frac{\sqrt{3}}{18} x \right)^3-8\left(\frac{\sqrt{3}}{18} x\right) + 8\right) = 0$$ Since $x>0$, we have $$\left( \frac{\sqrt{3}}{18} x \right)^3-8\left(\frac{\sqrt{3}}{18} x\right) + 8 = 0$$ Let $t=\frac{\sqrt{3}}{18} x$, we have $t^3-8t+8=0$. One can easily check that $t=2$ is a root, so we can factorize the cubic as such: $$(t-2)\left( t^2+2t-4 \right) = 0$$ The solutions of $t^2+2t-4=0$ are $t=-1+\sqrt{5}$ and $t=-1-\sqrt{5}$. However, since $x>0$, we have $t>0$ For $t=-1+\sqrt{5}$, it is positive, so it is valid. Obviously $t=-1-\sqrt{5}$ is negative, so it is rejected. Thus we have $t=2$ and $t=-1+\sqrt{5}$. Plugging $t=\frac{\sqrt{3}}{18} x$ and simplifying gives us our two solutions above. $\blacksquare$