$ABCD$ is a regular tetrahedron with side length 1. Points $X,$ $Y,$ and $Z,$ distinct from $A,$ $B,$ and $C,$ respectively, are drawn such that $BCDX,$ $ACDY,$ and $ABDZ$ are also regular tetrahedra. If the volume of the polyhedron with faces $ABC,$ $XYZ,$ $BXC,$ $XCY,$ $CYA,$ $YAZ,$ $AZB,$ and $ZBX$ can be written as $\frac{a\sqrt{b}}{c}$ for positive integers $a,b,c$ with $\gcd(a,c) = 1$ and $b$ squarefree, find $a+b+c.$

[i]Proposed by Jason Wang[/i]
1. **Define the coordinates of the tetrahedron:**
   Let the vertices of the regular tetrahedron \(ABCD\) be given in 4D coordinates as follows:
   \[
   D = (1,0,0,0), \quad A = (0,1,0,0), \quad B = (0,0,1,0), \quad C = (0,0,0,1)
   \]
   The side length of this tetrahedron is \(\sqrt{2}\), so we need to scale the volume by a factor of \(\frac{1}{2\sqrt{2}}\) at the end.

2. **Find the coordinates of \(X\):**
   Since \(X\) is the reflection of \(A\) over the centroid of \(\triangle BCD\), we first find the centroid \(G_x\) of \(\triangle BCD\):
   \[
   G_x = \left(\frac{1}{3}, 0, \frac{1}{3}, \frac{1}{3}\right)
   \]
   To find \(X\), we use the fact that \(\frac{X + A}{2} = G_x\):
   \[
   \frac{X + (0,1,0,0)}{2} = \left(\frac{1}{3}, 0, \frac{1}{3}, \frac{1}{3}\right)
   \]
   Solving for \(X\), we get:
   \[
   X = \left(\frac{2}{3}, -1, \frac{2}{3}, \frac{2}{3}\right)
   \]

3. **Find the coordinates of \(Y\) and \(Z\):**
   Similarly, we find the coordinates of \(Y\) and \(Z\):
   \[
   Y = \left(\frac{2}{3}, \frac{2}{3}, -1, \frac{2}{3}\right), \quad Z = \left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}, -1\right)
   \]

4. **Calculate the centroids of \(\triangle ABC\) and \(\triangle XYZ\):**
   The centroid \(G_1\) of \(\triangle ABC\) is:
   \[
   G_1 = \frac{A + B + C}{3} = \left(0, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)
   \]
   The centroid \(G_2\) of \(\triangle XYZ\) is:
   \[
   G_2 = \frac{X + Y + Z}{3} = \left(\frac{2}{3}, \frac{1}{9}, \frac{1}{9}, \frac{1}{9}\right)
   \]

5. **Calculate the distance between \(G_1\) and \(G_2\):**
   The distance \(h\) between \(G_1\) and \(G_2\) is:
   \[
   h = \sqrt{\left(\frac{2}{3}\right)^2 + 3 \left(\frac{2}{9}\right)^2} = \sqrt{\frac{4}{9} + \frac{4}{27}} = \sqrt{\frac{16}{27}} = \frac{4}{3\sqrt{3}}
   \]

6. **Calculate the side length of \(\triangle XYZ\):**
   The side length \(s_2\) of \(\triangle XYZ\) is:
   \[
   s_2 = \sqrt{\left(\frac{2}{3} - \frac{2}{3}\right)^2 + \left(-1 - \frac{2}{3}\right)^2 + \left(\frac{2}{3} - \frac{2}{3}\right)^2} = \frac{5}{3} \sqrt{2}
   \]

7. **Calculate the volume of the polyhedron:**
   The volume \(V\) of the polyhedron can be calculated using the formula for the volume of a frustum:
   \[
   V = \frac{h}{3} \left(\frac{B_1}{2} + \frac{B_2}{2} + \sqrt{B_1 B_2}\right)
   \]
   where \(B_1\) and \(B_2\) are the areas of the top and bottom hexagons, respectively. After scaling by \(\frac{1}{2\sqrt{2}}\), we get:
   \[
   V = \frac{128}{81} \times \frac{1}{\sqrt{8}} = \frac{32\sqrt{2}}{81}
   \]

The final answer is \(\boxed{32 + 2 + 81 = 115}\)