The seesaw is divided equally into $6$ parts and is already tilted to the left side with the first $2$ blocks. Assuming all the green blocks weigh the same, which side would the seesaw tilt if the $3^{\text{rd}}$ block is placed on the right edge of the seesaw, or would the seesaw not tilt at all?
Is this question even solvable? What is the proof?
You need to find the moment about the center hinge to decide, which side it would tilt. Let the weight of each of the green squares be $w$ and the distance between the blue pointers be $a$.
The two green squares are placed at a distance of $\dfrac{3a}2$ to the left of the hinge while one green square is placed at a distance of $3a$ to the right of the hinge.
The clockwise moment about the center hinge due to two green squares on the left is given by $2w \times \dfrac{3a}2 = 3aw$.
The counter-clockwise moment about the center hinge due to one green square on the right extreme is given by $w \times 3a = 3aw$.
Hence, the net clockwise moment about the hinge is $3aw - 3aw = 0$. Hence, the seesaw won't tilt.
The plots in the question and the solution were made using TikZ.