Problem 1510

### Part (a): Prove that the points $M, N, P, Q, R$ lie on the same plane.

1. **Inversion with Center $S$ and Power $SA^2$:**
- Given that $SA = 2a\sqrt{3}$, the power of inversion is $SA^2 = (2a\sqrt{3})^2 = 12a^2$.
- The points $M, N, P, Q, R$ are the projections of point $A$ on the lines $SB, SC, SD, SE, SF$ respectively.

2. Inversion Property:
- By the property of inversion, if $A$ is projected onto $M$ on line $SB$, then $SA^2 = SM \cdot SB$.
- Similarly, $SA^2 = SN \cdot SC = SP \cdot SD = SQ \cdot SE = SR \cdot SF$.

3. Coplanarity:
- Since $M, N, P, Q, R$ are images of $B, C, D, E, F$ under the inversion with center $S$ and power $SA^2$, they lie on the inverse $\omega'$ of the circumcircle of the hexagon $\omega$.
- The intersection of the inverse sphere of the plane $ABC$ and the inverse plane of the sphere through $S, \omega$ implies that $M, N, P, Q, R$ are coplanar.

Thus, the points $M, N, P, Q, R$ lie on the same plane.

### Part (b): Find the measure of the angle between the planes $(MNP)$ and $(ABC)$.

1. Tangent Planes:
- The tangent of $\omega$ at $A$ is also the tangent of $\omega'$ at $A$.
- Hence, the angle $\theta$ between the planes $ABC$ and $MNP$ is the angle formed by their diameters $AX$ and $AY$.

2. **Right Triangle $\triangle SAX$:**
- Consider $\triangle SAX$ where $Y$ is the projection of $A$ on $SX$.
- We have $SA = 2a\sqrt{3}$ and $AX = 2a$.

3. **Calculating $\theta$:**
- The angle $\theta$ is given by $\theta = \angle YAX = \angle ASX$.
- Using the tangent function, $\tan \theta = \frac{AX}{SA} = \frac{2a}{2a\sqrt{3}} = \frac{1}{\sqrt{3}}$.

4. **Finding $\theta$:**
- Therefore, $\theta = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = 30^\circ$.

The final answer is $\boxed{ \theta = 30^\circ }$.