Given the regular pyramid $SABC\ :\ A(\frac{\sqrt{3}a}{3},0,0),B(-\frac{\sqrt{3}a}{6},\frac{a}{2},0),C(-\frac{\sqrt{3}a}{6},-\frac{a}{2},0),S(0,0,\frac{\sqrt{3}a}{4})$, with $\ \triangle ABC$, side $a$ and satisfying $AM=2\cdot SO$. $O(0,0,0)$ and $M(-\frac{\sqrt{3}a}{6},0,0)$. Given $\vec{SA} = 25 \cdot \vec{SN} \Rightarrow N(\frac{\sqrt{3}a}{75},0,\frac{6\sqrt{3}a}{25})$. Point $P(0,0,\frac{2\sqrt{3}a}{9})$. Equation of the plane $(ABP)\ :\ 6x-4\sqrt{3}y+9z=2\sqrt{3}a$ with normal vector $(6,-4\sqrt{3},9)$. Equation of the plane $(SBC)\ :\ -3x+2z=\frac{\sqrt{3}a}{2}$ with normal vector $(-3,0,2)$. The normal vectors are perpendicular, so also the planes.