, 2022

Let the spheres with radii $11$, $13$, and $19$ have centers $P$, $Q$, and $R$, respectively, and let the three circles have common radius $r$. Segments $\overline{AP}$, $\overline{BQ}$, and $\overline{CR}$ are perpendicular to the plane of $\triangle ABC$, so $\overline{AB}$ is the projection of $\overline{PQ}$ onto that plane. Similarly, $\overline{AC}$ is the projection of $\overline{PR}$ onto that plane.

If $D$ is any point on the circle centered at $A$, then $\triangle PAD$ is a right triangle with $AD = r$ and $PD = 11$, so $AP^2 = 121 - r^2$. Similarly, $BQ^2 = 169 - r^2$ and $CR^2 = 361 - r^2$. The Pythagorean Theorem gives $PQ^2 = AB^2 + (BQ - AP)^2$, so from $PQ = 11 + 13 = 24$, it follows that
$$(BQ - AP)^2 = PQ^2 - AB^2 = 24^2 - 560 = 16.$$

Thus $BQ - AP = 4$ and $BQ^2 - AP^2 = (169 - r^2) - (121 - r^2) = 48$. Hence $BQ + AP = \frac{48}{4} = 12$, $BQ = 8$, $AP = 4$, and $r^2 = 105$.

Because $CR = \sqrt{19^2 - r^2} = 16$, it follows that
$$AC^2 = PR^2 - (CR - AP)^2 = (11 + 19)^2 - (16 - 4)^2 = 900 - 144 = 756.$$