To find the parametric equations of the line passing through the points $(1, 2, 3)$ and $(3, 2, 4)$, we can use the following approach:

Steps:

1. Direction Vector: The direction vector of the line is found by subtracting the coordinates of the two given points: $\mathbf{d} = (3 - 1, 2 - 2, 4 - 3) = (2, 0, 1)$

2. Parametric Equations: The parametric form of the line is given by: $x = x_0 + t \cdot d_x$ $y = y_0 + t \cdot d_y$ $z = z_0 + t \cdot d_z$ where $(x_0, y_0, z_0)$ is a point on the line (we can use $(1, 2, 3)$ as the starting point), and $(d_x, d_y, d_z)$ is the direction vector.

Thus, the parametric equations of the line are: $x = 1 + 2t, \quad y = 2 + 0t, \quad z = 3 + t$

Conclusion:

The correct answer is: $x = 1 + 2t, \quad y = 2, \quad z = 3 + t$

Let me know if you'd like further clarification or have any other questions!

Follow-up questions:

1. How do you find the direction vector between two points in 3D space?
2. What are the conditions for two parametric lines to be parallel?
3. How do you convert parametric equations into symmetric equations?
4. Can you determine the parametric equation of a line through two points in 2D?
5. How would you represent a line using a vector equation?

Tip:

When finding direction vectors, always subtract the coordinates of the second point from the first one to ensure correct direction!