Solution

Let's write the equilibrium condition for the two lower balls on the vertical and horizontal axes

$$\left\{\begin{matrix} mg = F_1 \sin\alpha\\ F_x = F_1 \cos\alpha& \end{matrix}\right.$$

And for the upper ball

$$T = mg + 2F_1 \sin\alpha$$

$$T = 3mg$$

Accordingly, when the thread burns out, the upper ball will be acted upon downwards by a force of $T=3mg$. From Newton's second law, we find its initial acceleration as

$$a = \frac{T}{m} = 3g$$

The lower balls will be acted upon in the horizontal direction by the force $F_x$, which will be compensated by the force $F_1 \cos\alpha$, and the force of gravity $mg$ — $F_1 \sin\alpha$

Thus, the lower balls will be in zero gravity

$$a=0$$

NO: Something similar happens when a Slinky falls

I recommend an interesting problem about Slinky IPhO 2019 "Springs and Slinky"