1. **Problem statement:** Find the angle between the surfaces given by $$xy^2z = 3x + z^2$$ and $$3x^2 - y^2 + 2z = 1$$ at the point $(1, -2, 1)$. 2. **Formula and approach:** The angle between two surfaces at a point is the angle between their normal vectors at that point. 3. **Find the gradients (normal vectors) of each surface:** For the first surface $F(x,y,z) = xy^2z - 3x - z^2 = 0$, $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ Calculate each partial derivative: $$\frac{\partial F}{\partial x} = y^2 z - 3$$ $$\frac{\partial F}{\partial y} = 2xy z$$ $$\frac{\partial F}{\partial z} = xy^2 - 2z$$ Evaluate at $(1, -2, 1)$: $$\frac{\partial F}{\partial x} = (-2)^2 \times 1 - 3 = 4 - 3 = 1$$ $$\frac{\partial F}{\partial y} = 2 \times 1 \times (-2) \times 1 = -4$$ $$\frac{\partial F}{\partial z} = 1 \times (-2)^2 - 2 \times 1 = 4 - 2 = 2$$ So, $$\nabla F(1,-2,1) = (1, -4, 2)$$ 4. For the second surface $G(x,y,z) = 3x^2 - y^2 + 2z - 1 = 0$, $$\nabla G = \left(\frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z}\right)$$ Calculate each partial derivative: $$\frac{\partial G}{\partial x} = 6x$$ $$\frac{\partial G}{\partial y} = -2y$$ $$\frac{\partial G}{\partial z} = 2$$ Evaluate at $(1, -2, 1)$: $$\frac{\partial G}{\partial x} = 6 \times 1 = 6$$ $$\frac{\partial G}{\partial y} = -2 \times (-2) = 4$$ $$\frac{\partial G}{\partial z} = 2$$ So, $$\nabla G(1,-2,1) = (6, 4, 2)$$ 5. **Calculate the angle $\theta$ between the two normal vectors:** $$\cos \theta = \frac{\nabla F \cdot \nabla G}{|\nabla F| |\nabla G|}$$ Calculate dot product: $$\nabla F \cdot \nabla G = 1 \times 6 + (-4) \times 4 + 2 \times 2 = 6 - 16 + 4 = -6$$ Calculate magnitudes: $$|\nabla F| = \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$$ $$|\nabla G| = \sqrt{6^2 + 4^2 + 2^2} = \sqrt{36 + 16 + 4} = \sqrt{56} = 2\sqrt{14}$$ 6. Substitute values: $$\cos \theta = \frac{-6}{\sqrt{21} \times 2\sqrt{14}} = \frac{-6}{2 \sqrt{294}} = \frac{-3}{\sqrt{294}}$$ Simplify $\sqrt{294}$: $$294 = 49 \times 6 = 7^2 \times 6$$ So, $$\cos \theta = \frac{-3}{7 \sqrt{6}}$$ 7. **Find the angle $\theta$:** $$\theta = \cos^{-1} \left( \frac{-3}{7 \sqrt{6}} \right)$$ This is the angle between the surfaces at the given point. **Final answer:** $$\boxed{\theta = \cos^{-1} \left( \frac{-3}{7 \sqrt{6}} \right)}$$