Consider what happens to the dot as the triangles rotate one-third of a turn: it must move through one-third of a circle of radius equal to the distance from the axle to the vertex. That radius is $6\sqrt3$, so the dot moves a circular distance of $4\pi\sqrt3$ units. It moves only $18$ linear units around the large triangle, however.
When the triangles rotate another third of a turn, the dot describes another circular arc, congruent to the first one, and moves another $18$ units linearly around the large triangle. It is now $12$ units from the lower lefthand vertex of the large triangle, so its distance from the lower lefthand axle is $\sqrt{3^2+\left(3\sqrt3\right)^2}=6$ units. On the next third of a turn of the triangles, therefore, it moves through a circular arc comprising one-third the circumference of a circle of radius $6$, so it moves a circular distance of $4\pi$ units. It still progresses $18$ units linearly around the large triangle, however, so it is now $6$ units from the lower lefthand vertex. The next third of a turn by the small triangles again moves it through a circular arc of length $4\pi\sqrt3$ units and takes it to the midpoint of the bottom edge of the large triangle. At this point it has moved a total of
$3\left(4\pi\sqrt3\right)+4\pi=4\pi\left(1+3\sqrt3\right)$
units.
The other half of its trip is a mirror image of the first half, so the total distance covered by the dot is $8\pi(1+3\sqrt3)$ units, or about $155.7263$ units.