Question 145, As shown in the figure, the side length of the equilateral triangle $\mathrm{ABC}$ is $4, P, Q$ are located on opposite sides of the plane of $\triangle \mathrm{ABC}$, satisfying $P A=P B=P C, Q A=Q B=Q C$, the dihedral angle $P-A B-Q$ is $120^{\circ}$, if there exists a point 0 such that the distances to $A, B, C, P,$ and $Q$ are all $\mathrm{d}$, then $\mathrm{d}=$ $\qquad$ -
Question 145, Solution: As shown in the figure, let the intersection point of $P Q$ with the plane of $\triangle A B C$ be $D$, then $D$ is the centroid of $\triangle A B C$. Taking the midpoint of $A B$ as $E$, point $D$ lies on $C E$, and $E D=\frac{2 \sqrt{3}}{3}, C D=\frac{4 \sqrt{3}}{3}$. By symmetry, point $O$ is clearly on $P Q$. Let $P D=x, Q D=y$, then $O D=\frac{|x-y|}{2}, d=\frac{x+y}{2}$. Since
$$
\begin{array}{l}
-\sqrt{3}=\tan (\angle P E D+\angle Q E D)=\frac{\tan \angle P E D+\tan \angle Q E D}{1-\tan \angle P E D \cdot \tan \angle Q E D}=\frac{\frac{P D, Q D}{E D} \frac{E D}{E D}}{1 \frac{P D D D D}{E D E D}}=\frac{\frac{x+y}{2 \sqrt{3}}}{1 \frac{x y}{4}} \\
\Rightarrow 3xy-4=2(x+y) \\
\end{array}
$$

Since $O P=O C=d$, we have
$$
\begin{array}{l}
O D^{2}+C D^{2}=d^{2} \\
\Rightarrow\left(\frac{x-y}{2}\right)^{2}+\left(\frac{4 \sqrt{3}}{3}\right)^{2}=\left(\frac{x-y}{2}\right)^{2} \\
\Rightarrow\left(\frac{x-y}{2}\right)^{2}+\left(\frac{4 \sqrt{3}}{3}\right)^{2}=\left(\frac{x-y}{2}\right)^{2} \\
\Rightarrow x y=\frac{16}{3}
\end{array}
$$

Substituting into (*) we get $x+y=6$, hence $d=\frac{x+y}{2}=3$.