[Geometry]
Problem: Spheres centered at points $P, Q, R$ are externally tangent to each other, and are tangent to plane $\mathcal{M}$ at points $P^{\prime}, Q^{\prime}, R^{\prime}$, respectively. All three spheres are on the same side of the plane. If $P^{\prime} Q^{\prime}=Q^{\prime} R^{\prime}=12$ and $P^{\prime} R^{\prime}=6$, compute the area of $\triangle P Q R$.

Solution: Let the radii be $p, q, r$ respectively. Looking at a cross-section of the spheres through $\overline{P Q}$ perpendicular to the plane, the points $P^{\prime}, P, Q, Q^{\prime}$ form a right trapezoid with $\overline{P^{\prime} P} \perp \overline{P^{\prime} Q^{\prime}}$ and $\overline{Q^{\prime} Q} \perp \overline{P^{\prime} Q^{\prime}}$. Draw $\overline{P M}$ perpendicular to $\overline{Q Q^{\prime}}$ as shown.

<img_3907>

Then $P P^{\prime}=M Q^{\prime}=p$ and $Q M=q-p$, while $P Q=p+q$ and $P M=P^{\prime} Q^{\prime}$. By the Pythagorean Theorem, $(q-p)^{2}+P^{\prime} Q^{\prime 2}=(p+q)^{2}$, so $q=\frac{\left(P^{\prime} Q^{\prime}\right)^{2}}{4 p}$. Thus $4 p q=P^{\prime} Q^{\prime 2}=12^{2}$. Similarly, $4 p r=P^{\prime} R^{\prime 2}=6^{2}$ and $4 q r=Q^{\prime} R^{\prime 2}=12^{2}$. Dividing the first equation by the third shows that $p=r$ (which can also be inferred from the symmetry of $\triangle P^{\prime} Q^{\prime} R^{\prime}$ ) and the equation $p r=9$ yields 3 as their common value; substitute in either of the other two equations to obtain $q=12$. Therefore the sides of $\triangle P Q R$ are $P Q=Q R=12+3=15$ and $P R=6$. The altitude to $\overline{P R}$ has length $\sqrt{15^{2}-3^{2}}=6 \sqrt{6}$, so the triangle's area is $\frac{1}{2}(6)(6 \sqrt{6})=\mathbf{1 8} \sqrt{\mathbf{6}}$.

Final answer: $18 \sqrt{6}$