Problem (Geometry):
Let $T=12$. Equilateral triangle $A B C$ is given with side length $T$. Points $D$ and $E$ are the midpoints of $\overline{A B}$ and $\overline{A C}$, respectively. Point $F$ lies in space such that $\triangle D E F$ is equilateral and $\triangle D E F$ lies in a plane perpendicular to the plane containing $\triangle A B C$. Compute the volume of tetrahedron $A B C F$.

Solution:
The volume of tetrahedron $A B C F$ is one-third the area of $\triangle A B C$ times the distance from $F$ to $\triangle A B C$. Since $D$ and $E$ are midpoints, $D E=\frac{B C}{2}=\frac{T}{2}$, and the distance from $F$ to $\triangle A B C$ is $\frac{T \sqrt{3}}{4}$. Thus the volume of $A B C F$ is $\frac{1}{3} \cdot \frac{T^{2} \sqrt{3}}{4} \cdot \frac{T \sqrt{3}}{4}=\frac{T^{3}}{16}$. With $T=12$, the answer is $\mathbf{1 0 8}$.

Final answer: 108