Problem 1253

Let $F$ stand for a value of the function $F$. Clearly, $x, y, z \in A$ iff the cubic $t^3 - 3t^2 + Ft - 1$ has three real roots. Normalise this to the form ($s = t - 1$)
$$s^3 - (3-F)s - (3-F) = 0.$$
If $a, b, c$ are the roots of this, then they are real iff
$$0 \le 4(3-F)^3 - 27(3-F)^2 = (F-3)^2(4(3-F) - 27) = -(F-3)^2(4F+15).$$
Consequently, the roots are real iff $F = 3$ or $F \le -15/4$. Now $F = 3$ means that $x, y, z$ satisfy the cubic equation $(t-1)^3 = 0$, and so $x = y = z = 1$.

Unless this occurs, then $F \leq -\frac{15}{4}$. If there is equality here, then two of $a, b, c$ are equal, i.e., two of $x, y, z$ are equal. Hence, $x = y = -1/2$, $z = 4$ say. Thus
$$F(A) = \left(-\infty, -\frac{15}{4}\right] \cup \{3\}.$$