Example 2 As shown in Figure 2, the three edges of the cube are $AB$, $BC$, and $CD$, and $AD$ is the body diagonal. Points $P$, $Q$, and $R$ are on $AB$, $BC$, and $CD$ respectively, with $AP=5$, $PB=15$, $BQ=15$, and $CR=10$. What is the area of the polygon formed by the intersection of the plane $PQR$ extended in all directions with the cube?
Explanation: Since $B P = B Q$, therefore, $P Q \parallel A C$. Thus, the line through point $R$ and parallel to $P Q$ intersects $A F$ at point $U$, and $A U = C R$.

Since the line through point $R$ and parallel to $P Q$ lies in the plane $P Q R$, $U$ is a vertex of the polygon formed by the intersection. Also, the midpoint of $U R$ is the center of the cube, so the points on the polygon formed by the intersection are symmetric about the center of the cube. Therefore, its area is twice the area of the trapezoid $P Q R U$.
It is easy to know that $U R = 20 \sqrt{2}, P Q = 15 \sqrt{2}, P U = 5 \sqrt{5}$.
Thus, the required area is
$$
\begin{array}{l}
(U R + P Q) \sqrt{P U^{2} - \left(\frac{U R - P Q}{2}\right)^{2}} \\
= 35 \times \sqrt{2} \times \sqrt{\frac{225}{2}} = 525 .
\end{array}
$$