1. **State the problem:** We need to find the directional derivative of the temperature function $$T = x^3 y + y^3 z + z^3 x$$ at the point $P = (2, -1, 0)$ in the direction towards the point $Q = (1, 1, 2)$. 2. **Formula for directional derivative:** The directional derivative of a scalar function $T(x,y,z)$ at point $P$ in the direction of a unit vector $\mathbf{u}$ is given by $$D_{\mathbf{u}} T = \nabla T(P) \cdot \mathbf{u}$$ where $\nabla T$ is the gradient vector of $T$. 3. **Find the gradient $\nabla T$:** Calculate partial derivatives: $$\frac{\partial T}{\partial x} = 3x^2 y + z^3$$ $$\frac{\partial T}{\partial y} = x^3 + 3y^2 z$$ $$\frac{\partial T}{\partial z} = y^3 + 3z^2 x$$ 4. **Evaluate $\nabla T$ at $P=(2,-1,0)$:** $$\frac{\partial T}{\partial x}(2,-1,0) = 3(2)^2(-1) + 0^3 = 3 \times 4 \times (-1) + 0 = -12$$ $$\frac{\partial T}{\partial y}(2,-1,0) = (2)^3 + 3(-1)^2 \times 0 = 8 + 0 = 8$$ $$\frac{\partial T}{\partial z}(2,-1,0) = (-1)^3 + 3 \times 0^2 \times 2 = -1 + 0 = -1$$ So, $$\nabla T(2,-1,0) = (-12, 8, -1)$$ 5. **Find the direction vector from $P$ to $Q$:** $$\mathbf{v} = Q - P = (1-2, 1 - (-1), 2 - 0) = (-1, 2, 2)$$ 6. **Normalize $\mathbf{v}$ to get unit vector $\mathbf{u}$:** $$|\mathbf{v}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$ $$\mathbf{u} = \frac{1}{3}(-1, 2, 2) = \left(-\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$$ 7. **Calculate the directional derivative:** $$D_{\mathbf{u}} T = \nabla T \cdot \mathbf{u} = (-12, 8, -1) \cdot \left(-\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right) = (-12)\left(-\frac{1}{3}\right) + 8 \left(\frac{2}{3}\right) + (-1) \left(\frac{2}{3}\right)$$ $$= 4 + \frac{16}{3} - \frac{2}{3} = 4 + \frac{14}{3} = \frac{12}{3} + \frac{14}{3} = \frac{26}{3}$$ **Final answer:** The directional derivative of $T$ at $(2,-1,0)$ towards $(1,1,2)$ is $$\boxed{\frac{26}{3}}$$.