After $4$ moves or $12$ steps the robot should have drawn something like a "flower".
It doesn't matter in which order $a, b, c$ are, since the trajectory forms a loop. A permutation of $a, b, c$ amounts to starting somewhere else on the loop or mirroring the loop. So wlog let's assume $a \le b \le c$.
Case 1.
If $ a+b \lt c $ then the loop forms something like an apple command key ⌘. It has $4$ revisited vertices (i.e. grid points).
Case 2.
If $ a+b = c $ then the path forms a "windmill" pattern with the loop crossing the center $4$ times. The path overlaps on the horizontal and vertical axes on a lengh $a$ in $4$ directions from the center.
That makes $4a+1$ revisited grid points.
Case 3.
If $ a+b \gt c $ with $a \lt b \lt c$ then the robot draws a four-fold knot pattern that crosses $8$ times. So $8$ revisited grid points.
Case 4.
If $ a+b \gt c $ with $a = b \lt c$ then it makes a square grid of $3 \times 3$ cells where the center segment of the borders is visited twice and has length $c - 2a$. That makes $4(c - 2a + 1)$ revisited grid points.
Case 5.
If ($ a+b \gt c) $ with $a \lt b = c$ then it forms something like a Swiss cross with a center square revisited.
That should be $4 \cdot a$ revisited grid points.
Case 6.
If ($ a+b \gt c $) with $a = b = c$ then we also have $4a$ revisited grid points. (but this time revisted and re-revisited)
I assume $a > 0$. This would be another special case with subcases.
