2. The base of the pyramid $M A B C$ is an isosceles $\triangle A B C$ ($A B \cong A C$). The foot of the height from the vertex $M$ is at the midpoint of the height $A A_{0}$ in $\triangle A B C$. A plane is placed through the edge $B C$ which is perpendicular to the edge $M A$ and intersects it at point $S$. It is known that this plane forms an angle of $30^{\circ}$ with the plane $A B C$. If the volume of the pyramid $S A B C$ is 2017, find the volume of the pyramid $M S B C$.
2. Let $N$ be the foot of the normal from $M$ to the plane $ABC$, which, according to the problem, is located at the midpoint of the segment $AA_0$. From the theorem of three normals, we have $MA_0 \perp BC$, so the entire plane $MA_0A$ is perpendicular to $BC$, and thus $SA_0 \perp BC$. Therefore, $\measuredangle SA_0A = 30^\circ$. Further, since $A_0S$ lies in the plane drawn through $BC$ defined in the problem, and this plane is perpendicular to $MA$, it follows that $A_0S \perp MA$. Since $\triangle SAA_0$ and $\triangle NAM$ are right triangles and share a common angle at vertex $A$, they are similar, so $\measuredangle NMA = \measuredangle SA_0A = 30^\circ$. Since $\triangle NMA$ is half of an equilateral triangle, it follows that $\triangle A_0MA$ is equilateral (since $N$ is the midpoint of side $A_0A$), and thus its height $A_0S$ is also its median, i.e., $SA \cong SM$. The pyramids $SABC$ and $MSBC$ share the common base $SBC$, and $SA$ and $SM$ are their heights to this base. Since these heights are congruent, the volumes of these two pyramids are equal. Therefore, the volume of the pyramid $MSBC$ is also 2017.

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