1. **State the problem:** Given that $\alpha + \beta = 2$ and the distance between points $(1, \alpha)$ and $(\beta, 1)$ is 3 units, find the value of $\alpha^2 + \beta^2$. 2. **Recall the distance formula:** The distance $d$ between points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Apply the distance formula:** Here, $$d = 3, \quad (x_1, y_1) = (1, \alpha), \quad (x_2, y_2) = (\beta, 1)$$ So, $$3 = \sqrt{(\beta - 1)^2 + (1 - \alpha)^2}$$ 4. **Square both sides to eliminate the square root:** $$9 = (\beta - 1)^2 + (1 - \alpha)^2$$ 5. **Expand the squares:** $$(\beta - 1)^2 = \beta^2 - 2\beta + 1$$ $$(1 - \alpha)^2 = 1 - 2\alpha + \alpha^2$$ 6. **Substitute back:** $$9 = \beta^2 - 2\beta + 1 + 1 - 2\alpha + \alpha^2$$ $$9 = \alpha^2 + \beta^2 - 2\alpha - 2\beta + 2$$ 7. **Group terms:** $$\alpha^2 + \beta^2 - 2(\alpha + \beta) + 2 = 9$$ 8. **Use the given sum $\alpha + \beta = 2$:** $$\alpha^2 + \beta^2 - 2(2) + 2 = 9$$ $$\alpha^2 + \beta^2 - 4 + 2 = 9$$ $$\alpha^2 + \beta^2 - 2 = 9$$ 9. **Solve for $\alpha^2 + \beta^2$:** $$\alpha^2 + \beta^2 = 9 + 2 = 11$$ **Final answer:** $$\boxed{11}$$