Three, as shown, in the triangular prism $A B C A_{1} B_{1} C_{1}$, all nine edges are equal to 1, and $\angle A_{1} A B$ $=\angle A_{1} A C$ $=\angle B A C$. Point $P$ is on the diagonal $A_{1} B$ of the side face $A_{1} A B B_{1}$, with $A_{1} P=\frac{\sqrt{3}}{3}$. Connect $P C_{1}$. Find the degree measure of the angle formed by the skew lines $P C_{1}$ and $A C$.
$$
\begin{array}{l} 
\text{ Given } A B=A C \\
= B C=1, \\
\angle B A C=60^{\circ} .
\end{array}
$$

So,
$$
\begin{array}{l}
\angle A_{1} A B \\
=\angle A_{1} A C \\
=\angle B A C=60^{\circ} .
\end{array}
$$

Thus, from $A_{1} A=A B=A C$ we know that $\triangle A_{1} A B$ and $\triangle \dot{A}_{1} A C$ are both equilateral triangles.
$\triangle A_{1} B C$ and $\triangle A_{1} B_{1} C_{1}$ are both equilateral triangles with side length 1.

Taking the midpoints of $B C$ and $B_{1} C_{1}$ as $M$ and $M_{1}$ respectively, and connecting $A_{1} M, A_{1} M_{1}, M_{1} M$, we have $B_{1} C_{1} \perp M M_{1}$. Thus, $\angle B_{1} C_{1} C=90^{\circ}$. Since $B B_{1}=B_{1} C_{1}=C_{1} C$ $=B C=1$, $B_{1} B C C_{1}$ is a square with side length 1. Therefore, $B C_{1}=\sqrt{2}$.

In $\triangle A_{1} B C_{1} \|_{1}$, we have $\angle B A_{1} C_{1}=90^{\circ}$. Therefore, in the right triangle $\triangle C_{1} A_{1} P$,
$$
\operatorname{tg} \angle A_{1} C_{1} P=\frac{A_{1} P}{A_{1} C_{1}}=\sqrt{3} .
$$

So, $\angle A_{1} C_{1} P=30^{\circ}$.