We are tasked with finding the parametric equations of the tangent line to the curve of intersection of two surfaces: the paraboloid $z = x^2 + y^2$ and the ellipsoid $3x^2 + 2y^2 + 2z^2 = 13$, at the point $(-1, 1, 2)$.

Steps to Solve:

1. Implicit differentiation: To find the direction of the tangent line at the point, we will first compute the gradients of both surfaces, as the tangent line to the curve of intersection is perpendicular to the gradients of both surfaces at the point.

2. Compute gradient of the paraboloid $z = x^2 + y^2$: $F_1(x, y, z) = z - (x^2 + y^2)$ The gradient of $F_1$ is: $\nabla F_1 = \left( \frac{\partial F_1}{\partial x}, \frac{\partial F_1}{\partial y}, \frac{\partial F_1}{\partial z} \right) = (-2x, -2y, 1)$ At the point $(-1, 1, 2)$, the gradient becomes: $\nabla F_1(-1, 1, 2) = (2, -2, 1)$

3. Compute gradient of the ellipsoid $3x^2 + 2y^2 + 2z^2 = 13$: $F_2(x, y, z) = 3x^2 + 2y^2 + 2z^2 - 13$ The gradient of $F_2$ is: $\nabla F_2 = \left( \frac{\partial F_2}{\partial x}, \frac{\partial F_2}{\partial y}, \frac{\partial F_2}{\partial z} \right) = (6x, 4y, 4z)$ At the point $(-1, 1, 2)$, the gradient becomes: $\nabla F_2(-1, 1, 2) = (-6, 4, 8)$

4. Find the direction of the tangent line: The tangent line is parallel to the cross product of the two gradients $\nabla F_1$ and $\nabla F_2$. We compute the cross product:

   \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ -6 & 4 & 8 \end{vmatrix}$$ Expanding the determinant: $$\mathbf{i} ((-2)(8) - (1)(4)) - \mathbf{j}((2)(8) - (1)(-6)) + \mathbf{k}((2)(4) - (-2)(-6))$$ $$= \mathbf{i}(-16 - 4) - \mathbf{j}(16 + 6) + \mathbf{k}(8 - 12)$$ $$= \mathbf{i}(-20) - \mathbf{j}(22) + \mathbf{k}(-4)$$ So, the cross product is: $$\nabla F_1 \times \nabla F_2 = (-20, -22, -4)$$

5. Parametric equations of the tangent line: The parametric equations for the tangent line take the form: $x(t) = x_0 + at, \quad y(t) = y_0 + bt, \quad z(t) = z_0 + ct$ where $(x_0, y_0, z_0)$ is the point $(-1, 1, 2)$, and $(a, b, c)$ is the direction vector $(-20, -22, -4)$.

   Therefore, the parametric equations are: $x(t) = -1 - 20t, \quad y(t) = 1 - 22t, \quad z(t) = 2 - 4t$

Final Answer:

The parametric equations of the tangent line are: $(x(t), y(t), z(t)) = (-1 - 20t, 1 - 22t, 2 - 4t)$

Would you like further clarifications, or do you have any additional questions?

Here are 5 related questions:

1. How is the gradient of a surface used in finding tangent lines?
2. Why is the cross product of two gradients used to find the direction of the tangent line?
3. Can you explain the geometric significance of the two surfaces intersecting?
4. How do parametric equations describe lines in 3D space?
5. What are other methods to compute tangent lines for more complex surfaces?

Tip: For parametric equations, the direction vector's components give insight into how fast each variable changes with respect to $t$.