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To find the vector equation for the tangent line at the point \( (1, 0, 1) \) where the two surfaces \( z = x^2 + y^2 + 2y \) and \( x + z = 2 \) intersect, we need to follow these steps:

### 1. Compute the gradient of each surface.
- For the surface \( f(x, y, z) = z - x^2 - y^2 - 2y = 0 \):
  \[
  \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (-2x, -2y - 2, 1)
  \]
  At the point \( (1, 0, 1) \), the gradient is:
  \[
  \nabla f(1, 0, 1) = (-2(1), -2(0) - 2, 1) = (-2, -2, 1)
  \]

- For the surface \( g(x, y, z) = x + z - 2 = 0 \):
  \[
  \nabla g = (1, 0, 1)
  \]
  At the point \( (1, 0, 1) \), the gradient is:
  \[
  \nabla g(1, 0, 1) = (1, 0, 1)
  \]

### 2. Find the direction of the tangent line.
The direction of the tangent line is given by the cross product of the two gradients \( \nabla f \) and \( \nabla g \):
\[
\vec{v} = \nabla f \times \nabla g
\]

Let's compute the cross product:
\[
\vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -2 & 1 \\ 1 & 0 & 1 \end{vmatrix}
\]
\[
\vec{v} = \mathbf{i} \begin{vmatrix} -2 & 1 \\ 0 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -2 & 1 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -2 & -2 \\ 1 & 0 \end{vmatrix}
\]
\[
\vec{v} = \mathbf{i}( -2 - 0 ) - \mathbf{j}( -2 - 1 ) + \mathbf{k}( 0 - (-2) )
\]
\[
\vec{v} = -2 \mathbf{i} + 3 \mathbf{j} + 2 \mathbf{k}
\]
Thus, the direction vector is \( \vec{v} = (-2, 3, 2) \).

### 3. Parametrize the tangent line.
The parametric equation of the tangent line is:
\[
\vec{r}(t) = (1, 0, 1) + t(-2, 3, 2)
\]
This simplifies to:
\[
\vec{r}(t) = (1 - 2t, 3t, 1 + 2t)
\]

### 4. Match with the options.
This matches the third option:
\[
\langle 1 + 2t, 3t, 1 - 2t \rangle
\]

So, the correct answer is:
\[
\boxed{\langle 1 + 2t, 3t, 1 - 2t \rangle}
\]

Would you like more details or further explanations on any step?

### Related Questions:
1. How do we interpret the gradient geometrically in the context of surfaces?
2. How do we calculate the cross product of two vectors in three dimensions?
3. What are the parametric equations of lines in vector form?
4. How can we verify the point of intersection of two surfaces?
5. What other methods exist to find tangent lines to curves formed by the intersection of surfaces?

### Tip:
Always ensure that the direction vector is correctly normalized or aligned if matching with multiple-choice answers, as direction matters but not necessarily the magnitude.

Find a vector equation for the tangent line to the curve of intersection of the surfaces z = x^2 + y^2 + 2y and x + z = 2 at the point (1, 0, 1).

This problem involves finding the vector equation of the tangent line to the curve of intersection between the surfaces z = x^2 + y^2 + 2y and x + z = 2 at the point (1, 0, 1). Learn how to compute the gradients of surfaces and apply the cross product to find the direction of the tangent line.

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