1. **State the problem:** We want to find the volume of the solid $D$ bounded above by the plane $z=3x+6y+24$, below by the plane $z=-1$, and laterally by the triangular region $T$ in the $xy$-plane with vertices $(2,2)$, $(4,0)$, and $(4,2)$. 2. **Describe the region $T$:** The triangle $T$ is given by $$T = \{(x,y) \in \mathbb{R}^2 : 2 \leq x \leq 4, \ -x+4 \leq y \leq 2 \}$$ 3. **Set up the volume integral:** The volume $V(D)$ is $$V(D) = \iiint_D 1 \, dV = \int_{x=2}^4 \int_{y=-x+4}^2 \int_{z=-1}^{3x+6y+24} 1 \, dz \, dy \, dx$$ 4. **Evaluate the inner integral over $z$:** $$\int_{-1}^{3x+6y+24} 1 \, dz = z \Big|_{-1}^{3x+6y+24} = (3x + 6y + 24) - (-1) = 3x + 6y + 25$$ 5. **Evaluate the middle integral over $y$:** $$\int_{-x+4}^2 (3x + 6y + 25) \, dy = \int_{-x+4}^2 (3x + 25) \, dy + \int_{-x+4}^2 6y \, dy$$ Calculate each part: $$ (3x + 25) y \Big|_{-x+4}^2 = (3x + 25)(2 - (-x + 4)) = (3x + 25)(x - 2) $$ $$ 3x + 25)(x - 2) = 3x^2 + 25x - 6x - 50 = 3x^2 + 19x - 50 $$ $$ \int_{-x+4}^2 6y \, dy = 3 y^2 \Big|_{-x+4}^2 = 3(2^2) - 3(-x+4)^2 = 12 - 3(x^2 - 8x + 16) = 12 - 3x^2 + 24x - 48 = -3x^2 + 24x - 36 $$ Sum these: $$3x^2 + 19x - 50 - 3x^2 + 24x - 36 = 43x - 86$$ 6. **Evaluate the outer integral over $x$:** $$\int_2^4 (43x - 86) \, dx = \left( \frac{43}{2} x^2 - 86x \right) \Big|_2^4 = \left( \frac{43}{2} \times 16 - 344 \right) - \left( \frac{43}{2} \times 4 - 172 \right)$$ Calculate each term: $$ \frac{43}{2} \times 16 = 43 \times 8 = 344 $$ $$ \frac{43}{2} \times 4 = 43 \times 2 = 86 $$ So, $$ (344 - 344) - (86 - 172) = 0 - (-86) = 86 $$ **Final answer:** $$\boxed{86}$$ The volume of the solid $D$ is 86.