The $\vec z$ that satisfy the two constraints are the points on the circle formed by the intersection of the unit sphere $\vec z^\mathrm T\vec z = 1$ and the plane $z_1 + z_2 + z_3 = 1$. Describe that circle using the parametric form explained at https://math.stackexchange.com/a/1184089/389981 : The circle passes through points $(1,0,0),(0,1,0),(0,0,1)$ which are evenly spaced on the circle, so its center is their average $(1,1,1)/3$ from which it follows that the radius is $\sqrt{2/3}$. By symmetry, the vector $(1,1,1)$ is normal to the plane in which the circle lies, so two orthogonal vectors in that plane are $(1,-1,0),(1,1,-2)$; normalized, they are $(1,-1,0)/\sqrt 2,(1,1,-2)/\sqrt 6$. Hence, a parametric description of points on the circle is\begin{align}(1,1,1)/3 & + \sqrt{2/3}\cos\theta\,(1,-1,0)/\sqrt 2\\
& + \sqrt{2/3}\sin\theta\,(1,1,-2)/\sqrt 6\end{align}
which simplifies to
$$(1,1,1)/3 +\cos\theta\,(1,-1,0)/\sqrt 3 +\sin\theta\,(1,1,-2)/3.$$
Now apply $\vec z^\mathrm T A\vec z$ and simplify to get
$$2(9 +\cos2\theta - 10\sin\theta + 2\sqrt3(\sin2\theta - \cos\theta))/9.$$
I used my graphing calculator to minimize and maximize that expression and found that the range of $\vec z^\mathrm T A\vec z$ is approximately
$[-0.529741,4.9184228]$ where the minimum and maximum values occur at approximate values of $\theta$ of $1.78286$ and $4.04074$.