$|x| = |y| = |z| = 1$ because points are on the unit circle.
Suppose, points $x,y$ $\text{and}$ $z$ makes angles $\theta_1,\theta_2$ $\text{and}$ $\theta_3$ respectively from +ve x-axis in anticlockwise direction.
$\therefore x = e^{i\theta_1}, y = e^{i\theta_2}, z = e^{i\theta_3}$
$x+y+z = 0$
$\frac{x+y+z}{z} = \frac{0}{z},z \neq 0 \Rightarrow \frac{x+y+z}{z} = 0$
$1+\frac{x}{z} + \frac{y}{z} = 0 \Rightarrow \frac{x}{z} + \frac{y}{z} = -1 \Rightarrow \left |\frac{x+y}{z} \right| = |-1| \Rightarrow \frac{|x+y|}{|z|}= 1 \Rightarrow |x+y| = 1$
$|e^{i\theta_1} + e^{i\theta_2}| = 1 \Rightarrow (\cos \theta_1 + \cos\theta_2)^2 + (\sin \theta_1 + \sin\theta_2)^2 = 1$
$ \Rightarrow \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 = \frac{-1}{2} \Rightarrow \cos(\theta_1 – \theta_2) = \frac{-1}{2}$
For $\theta \in [0, 360^{\circ}), |\theta_1 – \theta_2| = 120^{\circ}\; or\; |\theta_1 – \theta_2| = 240^{\circ}\;$
So, there are $2$ possibilities if we measure angle anticlockwise.
In the left figure, these 2 possibilities are shown for point $y$. The angle between lines $ox$ and $oy$ is $120^{\circ}$ and $240^{\circ}$.
We do the same thing for point $z$, so, divide by $y$ both sides in $x+y+z=0$ and do the same thing as above and we get the middle figure. Now, out of $2$ possibilities for each $y$ and $z$, if we fix $y$ then the other will be $z$, and suppose, we get the figure which is on the right side.
In the figure which is on the right side, since $ox=oy$, so angle between lines $oy$ & $yx$ and lines $ox$ & $yx$ will be same and it will be of $30^{\circ}$ because angle between $ox$ and $oy$ is $120^{\circ}$.
If we do the same for triangles $yoz$ and $zox$, we get all $3$ angles $\measuredangle zyx, \measuredangle yxz, \measuredangle xzy = 60^{\circ}.$
$\text{Hence,}$ $\textbf{Option (B)}$