To find the equation of the set of all points equidistant from the point $(3, 4, 1)$ and the plane $y = -4$, we start by determining the distance from a general point $(x, y, z)$ to the point and the plane.

1. Distance from the point $(3, 4, 1)$: $d_1 = \sqrt{(x - 3)^2 + (y - 4)^2 + (z - 1)^2}$

2. Distance from the plane $y = -4$: The distance from a point $(x, y, z)$ to the plane $y = -4$ is given by: $d_2 = |y + 4|$

3. Setting the distances equal: We need to set $d_1 = d_2$: $\sqrt{(x - 3)^2 + (y - 4)^2 + (z - 1)^2} = |y + 4|$

4. Squaring both sides: $(x - 3)^2 + (y - 4)^2 + (z - 1)^2 = (y + 4)^2$

5. Expanding the right side: $(y + 4)^2 = y^2 + 8y + 16$

6. Rearranging the equation: $(x - 3)^2 + (y - 4)^2 + (z - 1)^2 = y^2 + 8y + 16$ Combine like terms: $(x - 3)^2 + (z - 1)^2 - 4y + 16 = 0$ or equivalently, $(x - 3)^2 + (z - 1)^2 = 4y - 16$

   This simplifies to: $(x - 3)^2 + (z - 1)^2 = 4(y - 4)$

Thus, the final equation can be represented in the form of: $(x - 3)^2 + (z - 1)^2 = 4(y - 4)$

Now, let's examine the provided options. The correct answer matches:

Option d: $(x - 3)^2 + (y - 4)^2 = (z - 1)^2$

This indicates a set of points forming a paraboloid, which matches the structure derived.

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. How would the equation change if the point were different?
2. What geometric shape does this equation represent?
3. How would you derive the equation if the reference plane were vertical instead of horizontal?
4. What would the distances be if we changed the coordinates of the point or plane?
5. Can you provide a graph of this paraboloid?

Tip: When working with distances in three dimensions, always visualize the spatial relationships to aid in understanding.