1. **State the problem:** We have triangle ABC with points M on AB and K on AC. Given ratios: $\frac{AM}{MB} = \frac{3}{2}$ and $\frac{AK}{KC} = \frac{2}{3}$. We need to find the ratio of areas $\frac{S_{AKC}}{S_{ABC}}$. 2. **Recall area ratio rule:** The area of a triangle formed by a point dividing a side is proportional to the length ratio on that side. 3. **Analyze side AC:** Since $\frac{AK}{KC} = \frac{2}{3}$, point K divides AC into parts 2 and 3. The total length AC corresponds to 5 parts. 4. **Area ratio by base division:** The area of triangle AKC compared to ABC is proportional to the ratio of segment AC that AK covers. Since the height from B to AC is the same for both triangles, the ratio of areas equals the ratio of bases. 5. **Calculate area ratio:** $$\frac{S_{AKC}}{S_{ABC}} = \frac{AK}{AC} = \frac{2}{2+3} = \frac{2}{5}$$ 6. **Final answer:** $$\boxed{\frac{S_{AKC}}{S_{ABC}} = \frac{2}{5}}$$