Example 15 (Question from the 7th "Hope Cup" Invitational Competition) In the tetrahedron $P-ABC$, $\angle APC=\angle CPB=\angle BPA=\frac{\pi}{2}$, and $PA=PB=3, PC=4$. Also, $M$ is a point within the base $ABC$. The minimum value of the sum of the squares of the distances from $M$ to the three lateral faces of the tetrahedron is $\qquad$.
Solve: Fill in $\frac{144}{41}$. Reason: Since the three dihedral angles at the vertex of the tetrahedron are all $90^{\circ}$, this tetrahedron can be regarded as a corner of a rectangular prism, thus the property of a rectangular prism can be applied: the square of the diagonal equals the sum of the squares of the three sides.

As shown in Figure 7-14, let the distances from point $M$ to the three faces of the tetrahedron be $M D$, $M E$, and $M F$, then $M D^{2}+M E^{2}+M F^{2}=M P^{2}$. When $M P \perp$ plane $A B C$, $M P$ is the shortest, at this time the tetrahedrons $P-A B C$ and $A-P B C$ have equal volumes, so $M P=\frac{P A \cdot P B \cdot P C}{2 S_{\triangle A B C}}$, where $P A=$ $P B=3, P C=4, A C=B C=5, A B=3 \sqrt{2}, S_{\triangle A B C}=$ $\frac{3}{2} \sqrt{41}$. Therefore, $M P=\frac{12}{\sqrt{41}}$, which means $M P^{2}=\frac{144}{41}$.