To solve the given problems step by step, we will follow the mathematical principles of distance in three-dimensional geometry.
### Part (i): Finding Points on the X-axis
1. **Identify the point and distance**: We have the point \( B(1, -2, 3) \) and we need to find points \( A(x, 0, 0) \) on the X-axis that are at a distance of \( 2\sqrt{6} \) from point \( B \).
2. **Distance formula**: The distance \( d \) between two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) in 3D space is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
3. **Set up the equation**: Using the distance formula, we have:
\[
2\sqrt{6} = \sqrt{(1 - x)^2 + (-2 - 0)^2 + (3 - 0)^2}
\]
4. **Square both sides**: To eliminate the square root, we square both sides:
\[
(2\sqrt{6})^2 = (1 - x)^2 + (-2)^2 + (3)^2
\]
This simplifies to:
\[
24 = (1 - x)^2 + 4 + 9
\]
\[
24 = (1 - x)^2 + 13
\]
5. **Rearranging the equation**: Subtract 13 from both sides:
\[
11 = (1 - x)^2
\]
6. **Taking the square root**: Taking the square root gives:
\[
1 - x = \pm \sqrt{11}
\]
7. **Solving for x**: This leads to two equations:
- \( 1 - x = \sqrt{11} \) → \( x = 1 - \sqrt{11} \)
- \( 1 - x = -\sqrt{11} \) → \( x = 1 + \sqrt{11} \)
8. **Final points**: Therefore, the points on the X-axis are:
\[
A_1(1 - \sqrt{11}, 0, 0) \quad \text{and} \quad A_2(1 + \sqrt{11}, 0, 0)
\]
### Part (ii): Finding Points on the Y-axis
1. **Identify the point and distance**: We have the point \( B(3, -2, 5) \) and we need to find points \( A(0, y, 0) \) on the Y-axis that are at a distance of \( 5\sqrt{2} \) from point \( B \).
2. **Set up the equation**: Using the distance formula:
\[
5\sqrt{2} = \sqrt{(3 - 0)^2 + (-2 - y)^2 + (5 - 0)^2}
\]
3. **Square both sides**: Squaring both sides gives:
\[
(5\sqrt{2})^2 = (3 - 0)^2 + (-2 - y)^2 + (5 - 0)^2
\]
This simplifies to:
\[
50 = 9 + (-2 - y)^2 + 25
\]
\[
50 = 34 + (-2 - y)^2
\]
4. **Rearranging the equation**: Subtract 34 from both sides:
\[
16 = (-2 - y)^2
\]
5. **Taking the square root**: Taking the square root gives:
\[
-2 - y = \pm 4
\]
6. **Solving for y**: This leads to two equations:
- \( -2 - y = 4 \) → \( y = -6 \)
- \( -2 - y = -4 \) → \( y = 2 \)
7. **Final points**: Therefore, the points on the Y-axis are:
\[
A_1(0, -6, 0) \quad \text{and} \quad A_2(0, 2, 0)
\]
### Summary of Solutions
- Points on the X-axis: \( (1 - \sqrt{11}, 0, 0) \) and \( (1 + \sqrt{11}, 0, 0) \)
- Points on the Y-axis: \( (0, -6, 0) \) and \( (0, 2, 0) \)
