1. **Problem Statement:** Find the total electric potential energy of a system composed of three charges: $q_1 = 10q$, $q_2 = -5q$, and $q_3 = 8q$, located at the vertices of a triangle with sides $3d$, $4d$, and $5d$. 2. **Formula:** The total electric potential energy $U$ of a system of point charges is given by the sum of the potential energies between each pair: $$U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right)$$ where $k$ is Coulomb's constant, and $r_{ij}$ is the distance between charges $q_i$ and $q_j$. 3. **Identify distances:** - Between $q_1$ and $q_2$: $r_{12} = 3d$ - Between $q_1$ and $q_3$: $r_{13} = 4d$ - Between $q_2$ and $q_3$: $r_{23} = 5d$ 4. **Substitute values:** $$U = k \left( \frac{10q \times (-5q)}{3d} + \frac{10q \times 8q}{4d} + \frac{-5q \times 8q}{5d} \right)$$ 5. **Calculate each term:** $$U = k \left( \frac{-50q^2}{3d} + \frac{80q^2}{4d} + \frac{-40q^2}{5d} \right)$$ 6. **Simplify fractions:** $$U = k \left( -\frac{50q^2}{3d} + \frac{80q^2}{4d} - \frac{40q^2}{5d} \right)$$ 7. **Find common denominator $60d$ to combine terms:** $$-\frac{50q^2}{3d} = -\frac{50q^2 \times 20}{60d} = -\frac{1000q^2}{60d}$$ $$\frac{80q^2}{4d} = \frac{80q^2 \times 15}{60d} = \frac{1200q^2}{60d}$$ $$-\frac{40q^2}{5d} = -\frac{40q^2 \times 12}{60d} = -\frac{480q^2}{60d}$$ 8. **Sum the numerators:** $$-1000q^2 + 1200q^2 - 480q^2 = (-1000 + 1200 - 480)q^2 = (-280)q^2$$ 9. **Final expression:** $$U = k \times \frac{-280q^2}{60d}$$ 10. **Simplify fraction:** $$\frac{-280}{60} = \frac{\cancel{-280}}{\cancel{60}} = \frac{-14}{3}$$ 11. **Answer:** $$\boxed{U = -\frac{14}{3} \frac{k q^2}{d}}$$ This means the total electric potential energy of the system is negative, indicating an overall attractive interaction.