To solve the problem step by step, we will determine the locus of the centroid of triangle ABC formed by the intersection of a variable plane with the coordinate axes.
### Step 1: Define the equation of the plane
Let the equation of the plane be given by:
\[ Ax + By + Cz + D = 0 \]
### Step 2: Use the distance formula
The distance \( k \) from the origin (0, 0, 0) to the plane is given by the formula:
\[ \text{Distance} = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \]
Since the distance is given as \( k \), we can set up the equation:
\[ k = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \]
### Step 3: Express \( D \) in terms of \( k \)
From the distance formula, we can rearrange to find \( D \):
\[ |D| = k \sqrt{A^2 + B^2 + C^2} \]
Thus, we can write:
\[ D = \pm k \sqrt{A^2 + B^2 + C^2} \]
For simplicity, we can consider \( D = k \sqrt{A^2 + B^2 + C^2} \).
### Step 4: Find the coordinates of points A, B, and C
The points where the plane intersects the axes can be found by setting two variables to zero in the plane equation.
1. **Point A (x-intercept)**: Set \( y = 0 \) and \( z = 0 \):
\[ Ax + D = 0 \Rightarrow x = -\frac{D}{A} \]
Thus, \( A = \left(-\frac{D}{A}, 0, 0\right) \).
2. **Point B (y-intercept)**: Set \( x = 0 \) and \( z = 0 \):
\[ By + D = 0 \Rightarrow y = -\frac{D}{B} \]
Thus, \( B = \left(0, -\frac{D}{B}, 0\right) \).
3. **Point C (z-intercept)**: Set \( x = 0 \) and \( y = 0 \):
\[ Cz + D = 0 \Rightarrow z = -\frac{D}{C} \]
Thus, \( C = \left(0, 0, -\frac{D}{C}\right) \).
### Step 5: Coordinates of the centroid of triangle ABC
The coordinates of the centroid \( G \) of triangle ABC can be calculated as:
\[
G\left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}, \frac{z_A + z_B + z_C}{3} \right)
\]
Substituting the coordinates of A, B, and C:
\[
G\left( \frac{-\frac{D}{A} + 0 + 0}{3}, \frac{0 - \frac{D}{B} + 0}{3}, \frac{0 + 0 - \frac{D}{C}}{3} \right) = \left( -\frac{D}{3A}, -\frac{D}{3B}, -\frac{D}{3C} \right)
\]
### Step 6: Substitute \( D \) and find the locus
Substituting \( D = k \sqrt{A^2 + B^2 + C^2} \):
\[
G\left( -\frac{k \sqrt{A^2 + B^2 + C^2}}{3A}, -\frac{k \sqrt{A^2 + B^2 + C^2}}{3B}, -\frac{k \sqrt{A^2 + B^2 + C^2}}{3C} \right)
\]
Let:
\[
x = -\frac{k \sqrt{A^2 + B^2 + C^2}}{3A}, \quad y = -\frac{k \sqrt{A^2 + B^2 + C^2}}{3B}, \quad z = -\frac{k \sqrt{A^2 + B^2 + C^2}}{3C}
\]
### Step 7: Relate \( A, B, C \) to \( x, y, z \)
From the expressions for \( x, y, z \):
\[
A = -\frac{k \sqrt{A^2 + B^2 + C^2}}{3x}, \quad B = -\frac{k \sqrt{A^2 + B^2 + C^2}}{3y}, \quad C = -\frac{k \sqrt{A^2 + B^2 + C^2}}{3z}
\]
### Step 8: Substitute back into the distance equation
Substituting these values into the distance equation gives:
\[
k^2 = D^2 \left( \frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} \right)
\]
After simplification, we arrive at:
\[
\frac{9}{k^2} = \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}
\]
### Final Result
The locus of the centroid of triangle ABC is given by:
\[
\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{k^2}
\]
