1. Identify the given elements and their properties:
- Triangle $ABC$ with sides $AB = 26$, $AC = 28$, and $BC = 30$.
- $X$, $Y$, and $Z$ are the midpoints of arcs $BC$, $CA$, and $AB$ (not containing the opposite vertices) respectively on the circumcircle of $\triangle ABC$.
- $P$ is the midpoint of arc $BC$ containing point $A$.
- Lines $BP$ and $XZ$ meet at $M$, and lines $CP$ and $XY$ meet at $N$.

2. **Determine the properties of points $X$, $Y$, $Z$, and $P$:**
- Since $X$ is the midpoint of arc $BC$ not containing $A$, $X$ is equidistant from $B$ and $C$, i.e., $XB = XC$.
- Similarly, $Y$ and $Z$ are equidistant from their respective arc endpoints.
- $P$ is the midpoint of arc $BC$ containing $A$, so $P$ is also equidistant from $B$ and $C$.

3. Use angle chasing to find relationships between angles:
- Since $X$ is the midpoint of arc $BC$, $\angle BXC = 180^\circ - \angle BAC$.
- Similarly, $\angle YZA = 180^\circ - \angle ABC$ and $\angle ZXY = 180^\circ - \angle ACB$.

4. **Fold lines $XB$ and $XC$ over $XM$ and $XN$:**
- Since $XB = XC$ and $X$ is the midpoint of arc $BC$, folding $XB$ over $XM$ and $XC$ over $XN$ will make $B$ and $C$ coincide.
- This implies that $M$ and $N$ are reflections of $B$ and $C$ over $X$.

5. **Determine the perpendicularity of $XB$ and $XC$ to $MN$:**
- Since $X$ and $P$ are diametrically opposite, $\angle XBM = \angle XCN = 90^\circ$.
- Therefore, $XB$ and $XC$ are perpendicular to $MN$.

6. **Calculate the distance from $X$ to $MN$:**
- The distance from $X$ to $MN$ is equal to $XC$, which is the radius of the circumcircle of $\triangle ABC$.

7. **Use the circumradius-area formula to find the circumradius $R$:**
- The area $K$ of $\triangle ABC$ can be found using Heron's formula:
$$s = \frac{AB + AC + BC}{2} = \frac{26 + 28 + 30}{2} = 42$$
$$K = \sqrt{s(s - AB)(s - AC)(s - BC)} = \sqrt{42 \cdot (42 - 26) \cdot (42 - 28) \cdot (42 - 30)} = \sqrt{42 \cdot 16 \cdot 14 \cdot 12} = \sqrt{112896} = 336$$
- The circumradius $R$ is given by:
$$R = \frac{abc}{4K} = \frac{26 \cdot 28 \cdot 30}{4 \cdot 336} = \frac{21840}{1344} = 16.25$$
- Therefore, the distance from $X$ to $MN$ is $R = 16.25$.

8. Square the distance to find the final answer:
$$(16.25)^2 = 264.0625$$

The final answer is $\boxed{264.0625}$