Let $A B C D$ be a regular tetrahedron, and let $P$ be an interior point of the face $A B C$. We know the lengths of $P A, P B, P C$. Calculate the length of the segment $P D$!
Let the lengths of $P A, P B, P C$ be denoted by $a$, $b$, and $c$ respectively. Rotate the triangle $A B P$ around $A$ such that $B$ coincides with $C$, and let the new position of $P$ be $Q$. Since triangle $A B C$ is equilateral, triangle $A P Q$ is also equilateral. Thus, the lengths of the sides of triangle $C P Q$ are exactly the three given distances. This implies that the problem is only solvable if the sides $a, b, c$ can form a triangle. Furthermore, the triangle with sides $P Q=a, Q C=b$, and $C P=c$ has the side $P Q$ extended outward to form the equilateral triangle $A P Q$, and the points $A$ and $C$ determine the side $A C$ of the equilateral triangle $A B C$. Let this side be denoted by $d$. Applying the cosine rule to triangles $A P C$ and $C P Q$, we get

\[
\begin{aligned}
d^{2} & =a^{2}+c^{2}-2 a c \cos \left(60^{\circ}+\beta\right)=a^{2}+c^{2}-2 a c\left(\frac{1}{2} \cos \beta-\frac{\sqrt{3}}{2} \sin \beta\right)= \\
& =a^{2}+c^{2}-\frac{1}{2}\left(a^{2}+c^{2}-b^{2}\right)+\sqrt{3} a c \sin \beta=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)+2 \sqrt{3} t
\end{aligned}
\]

where $\beta$ is the angle of triangle $C P Q$, and $t$ is the area of triangle $C P Q$.

Let $D$ be the projection of $P$ onto the plane of triangle $A B C$, and let $K$ be the point of projection. By the Pythagorean theorem, $P D^{2}=P K^{2}+K D^{2}$. This can be easily determined based on $d$, since if the center of the tetrahedron is denoted by $O$, then $O$ divides $D K$ into quarters, and $A O=O D$. Therefore, in triangle $A K O$,

\[
\left(\frac{3}{4} D K\right)^{2}=\left(\frac{1}{4} D K\right)^{2}+A K^{2}
\]

so $D K^{2}=2 A K^{2}=\frac{2}{3} d^{2}$. We will show that

\[
P K^{2}=\frac{a^{2}+b^{2}+c^{2}}{3}-A K^{2}
\]

from which we get

\[
\begin{gathered}
P D^{2}=\frac{a^{2}+b^{2}+c^{2}}{3}-\frac{d^{2}}{3}+\frac{2 d^{2}}{3}=\frac{1}{3}\left(a^{2}+b^{2}+c^{2}+d^{2}\right)= \\
=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)+\frac{2 t}{\sqrt{3}}
\end{gathered}
\]

It is known that for a triangle with sides $a, b, c$,

\[
16 t^{2}=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)
\]

so

\[
P D^{2}=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)+\frac{1}{2 \sqrt{3}} \sqrt{2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)}
\]

We still need to verify the (1) relationship. Choose the origin of the coordinate system at point $K$, and denote the coordinates of the points in question by the corresponding lowercase letters with single and double indices. Thus, we get

\[
\begin{gathered}
P K^{2}=p_{1}^{2}+p_{2}^{2} \\
a^{2}=\left(a_{1}-p_{1}\right)^{2}+\left(a_{2}-p_{2}\right)^{2} \\
b^{2}=\left(b_{1}-p_{1}\right)^{2}+\left(b_{2}-p_{2}\right)^{2} \\
c^{2}=\left(c_{1}-p_{1}\right)^{2}+\left(c_{2}-p_{2}\right)^{2} \\
A K^{2}=a_{1}^{2}+a_{2}^{2}
\end{gathered}
\]

Substituting these into (1) and considering that $A K^{2}=B K^{2}=C K^{2}, a_{1}+b_{1}+c_{1}=0, a_{2}+b_{2}+c_{2}=0$, we indeed get an identity. The last two relationships are true because $K$ is not only the center of the circumcircle of triangle $A B C$, but also its centroid.

Remark. The relationship between the segments $a, b, c, d$ can be brought into a symmetric form:

\[
\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}=a^{4}+b^{4}+c^{4}+d^{4}
\]