Given points $A(1,1,1)$ and $P(1,1,0)$ in the spatial coordinate system. Rotate the point $P$ by $60^{\circ}$ in the positive direction around the ray $O A$. Determine the coordinates of the rotated point.
In the cube, where $OA$ is the body diagonal, $P$ is one of the vertices adjacent to $A$ by an edge. Then, the other two vertices adjacent to $A$ from the positive direction are $Q(0; 1; 1)$ and $R(1; 0; 1)$ (Figure 1). Thus, the triangle $PQR$ is equilateral. The positive $120^{\circ}$ rotation around the $OA$ axis carries the cube, and thus the triangle $PQR$, onto itself. When the point $P$ is rotated around the $OA$ axis, it rotates in the plane of the triangle $PQR$ around the intersection point $C$ of the axis and the plane (Figure 2).

Therefore, we need to rotate the point $P$ by a positive $60^{\circ}$ around $C$ in the plane of $PQR$. For this, we need the coordinates of the point $C$, which can be determined in several ways.

In the cube, $\overrightarrow{AP} + \overrightarrow{AQ} + \overrightarrow{AR} = \overrightarrow{AO}$, and if $S$ is the centroid of the triangle $PQR$, it is known that $\overrightarrow{AP} + \overrightarrow{AQ} + \overrightarrow{AR} = 3 \overrightarrow{AS}$. Thus, for the point $S$ in the plane of $PQR$, $3 \overrightarrow{AS} = \overrightarrow{AO}$, and the plane of $PQR$ intersects the segment $OA$ at its point farthest from $O$: $C = S\left(\frac{2}{3}; \frac{2}{3}; \frac{2}{3}\right)$.

The positive $60^{\circ}$ rotation around $C$ can be represented as the composition of a negative $120^{\circ}$ rotation around $C$ and a positive $180^{\circ}$ rotation, which is a reflection in $C$. Thus, the desired point $P'$ is the point obtained by a negative $120^{\circ}$ rotation of $P$ around $C$, which is the reflection of $R$ in $C$ (Figure 3).

Using the appropriate position vectors, $\mathbf{c} = \frac{\mathbf{p}' + \mathbf{r}}{2}$, from which $\mathbf{p}' = 2 \mathbf{c} - \mathbf{r}$. The coordinates of $P'$ are thus $\frac{4}{3} - 1; \frac{4}{3}; \frac{4}{3} - 1$, that is, $P'\left(\frac{1}{3}; \frac{4}{3}; \frac{1}{3}\right)$.

Based on multiple solutions

Note: The positive rotation direction around the $OA$ half-line is to be understood from the half-space into which the half-line points.