1. **State the problem:** Evaluate the triple integral $$\int_{-1}^2 \int_1^{y+2} \int_e^{2e} \frac{(x+y)}{z} \, dz \, dx \, dy.$$ 2. **Inner integral:** Integrate with respect to $z$ first. The integral is $$\int_e^{2e} \frac{(x+y)}{z} \, dz.$$ Since $x$ and $y$ are constants with respect to $z$, factor them out: $$ (x+y) \int_e^{2e} \frac{1}{z} \, dz = (x+y) [\ln z]_e^{2e} = (x+y)(\ln(2e) - \ln e) = (x+y) \ln 2.$$ 3. **Middle integral:** Now integrate with respect to $x$: $$\int_1^{y+2} (x+y) \ln 2 \, dx = \ln 2 \int_1^{y+2} (x+y) \, dx.$$ Calculate the integral inside: $$\int_1^{y+2} (x+y) \, dx = \left[ \frac{x^2}{2} + yx \right]_1^{y+2} = \left( \frac{(y+2)^2}{2} + y(y+2) \right) - \left( \frac{1}{2} + y \right).$$ Simplify: $$\frac{(y+2)^2}{2} + y(y+2) - \frac{1}{2} - y = \frac{y^2 + 4y + 4}{2} + y^2 + 2y - \frac{1}{2} - y = \frac{y^2}{2} + 2y + 2 + y^2 + 2y - \frac{1}{2} - y = \frac{3y^2}{2} + 3y + \frac{3}{2}.$$ So the middle integral is $$\ln 2 \left( \frac{3y^2}{2} + 3y + \frac{3}{2} \right) = \frac{3 \ln 2}{2} (y^2 + 2y + 1) = \frac{3 \ln 2}{2} (y+1)^2.$$ 4. **Outer integral:** Integrate with respect to $y$: $$\int_{-1}^2 \frac{3 \ln 2}{2} (y+1)^2 \, dy = \frac{3 \ln 2}{2} \int_{-1}^2 (y+1)^2 \, dy.$$ Calculate the integral inside: $$\int_{-1}^2 (y+1)^2 \, dy = \left[ \frac{(y+1)^3}{3} \right]_{-1}^2 = \frac{(2+1)^3}{3} - \frac{( -1 + 1)^3}{3} = \frac{3^3}{3} - 0 = \frac{27}{3} = 9.$$ 5. **Final answer:** Multiply: $$\frac{3 \ln 2}{2} \times 9 = \frac{27 \ln 2}{2}.$$ **Answer:** $$\boxed{\frac{27}{2} \ln 2}.$$