Let $PA_1A_2...A_{12} $  be the regular pyramid, $ A_1A_2...A_{12} $ is regular polygon,  $S$ is area of the triangle $PA_1A_5$ and angle  between of the planes $A_1A_2...A_{12} $ and $ PA_1A_5 $ is equal to $ \alpha $.
Find the volume of the pyramid.
1. **Define Variables and Relationships:**
   - Let \( H \) be the height of the pyramid.
   - Let \( R \) be the radius of the circumcircle of the base.
   - Let \( h \) be the height of the triangle \( PA_1A_5 \).
   - Let \( a \) be the length of the segment \( A_1A_5 \).
   - Let \( A \) be the midpoint of \( A_1A_5 \).
   - Let \( O \) be the center of the base.
   - Let \( S_{base} \) be the area of the base.

2. **Volume of the Pyramid:**
   The volume \( V \) of the pyramid is given by:
   \[
   V = \frac{1}{3} S_{base} H
   \]

3. **Area of Triangle \( PA_1A_5 \):**
   The area \( S \) of the triangle \( PA_1A_5 \) is:
   \[
   S = \frac{1}{2} a h
   \]

4. **Height Relationship:**
   From the triangle \( SOA \), we have:
   \[
   H = h \sin \alpha
   \]
   and
   \[
   AO = \frac{R}{2} = h \cos \alpha
   \]

5. **Area of the Base:**
   The area of the base \( S_{base} \) is:
   \[
   S_{base} = \frac{(2h \cos \alpha)^2 \sin \frac{\pi}{3}}{2} = (h \cos \alpha)^2 \sin \frac{\pi}{3}
   \]
   Since \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \), we get:
   \[
   S_{base} = (h \cos \alpha)^2 \cdot \frac{\sqrt{3}}{2}
   \]

6. **Length of Segment \( A_1A_5 \):**
   From the triangle \( A_1AO \), we have:
   \[
   AO = \frac{R}{2} = \frac{A_1A_5}{2 \sqrt{3}} = \frac{a}{4 \sqrt{3}} = h \cos \alpha
   \]
   Hence,
   \[
   h = \frac{a}{4 \sqrt{3} \cos \alpha}
   \]

7. **Height \( h \) in Terms of \( S \):**
   Using the area \( S \) of the triangle \( PA_1A_5 \):
   \[
   S = \frac{1}{2} a h \implies h = \frac{2S}{a}
   \]
   Substituting \( a = 4 \sqrt{3} h \cos \alpha \):
   \[
   h = \frac{2S}{4 \sqrt{3} h \cos \alpha} \implies h^2 = \frac{S}{2 \sqrt{3} \cos \alpha} \implies h = \sqrt{\frac{S}{2 \sqrt{3} \cos \alpha}}
   \]

8. **Volume Calculation:**
   Using the above results, the volume \( V \) is:
   \[
   V = \frac{1}{3} (h \cos \alpha)^2 h \sin \alpha = \frac{1}{3} h^3 \cos^2 \alpha \sin \alpha
   \]
   Substituting \( h = \sqrt{\frac{S}{2 \sqrt{3} \cos \alpha}} \):
   \[
   V = \frac{1}{3} \left( \sqrt{\frac{S}{2 \sqrt{3} \cos \alpha}} \cos \alpha \right)^2 \sqrt{\frac{S}{2 \sqrt{3} \cos \alpha}} \sin \alpha
   \]
   Simplifying:
   \[
   V = \frac{1}{3} \left( \frac{S \cos \alpha}{2 \sqrt{3} \cos \alpha} \right) \sqrt{\frac{S}{2 \sqrt{3} \cos \alpha}} \sin \alpha
   \]
   \[
   V = \frac{1}{3} \left( \frac{S}{2 \sqrt{3}} \right) \sqrt{\frac{S}{2 \sqrt{3} \cos \alpha}} \sin \alpha
   \]
   \[
   V = \frac{S \sqrt{S} \sqrt{\cos \alpha} \sin \alpha}{2 \sqrt{2} \sqrt[4]{27}}
   \]

The final answer is \( \boxed{ \frac{S \sqrt{S} \sqrt{\cos \alpha} \sin \alpha}{2 \sqrt{2} \sqrt[4]{27}} } \)