To solve the problem, we need to find the locus of the centroid of triangle ABC formed by the intersection of a variable plane with the coordinate axes, given that the plane is at a distance of 3 units from the origin.
### Step-by-Step Solution:
1. **Equation of the Plane**:
The equation of a plane in intercept form is given by:
\[
\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1
\]
where A, B, and C are the x, y, and z-intercepts of the plane respectively.
2. **Distance from the Origin**:
The distance \(d\) from the origin to the plane can be calculated using the formula:
\[
d = \frac{|1|}{\sqrt{\left(\frac{1}{A}\right)^2 + \left(\frac{1}{B}\right)^2 + \left(\frac{1}{C}\right)^2}} = 3
\]
Squaring both sides, we have:
\[
1 = 9 \left(\frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2}\right)
\]
This simplifies to:
\[
\frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2} = \frac{1}{9}
\]
3. **Centroid of Triangle ABC**:
The points A, B, and C where the plane intersects the axes are:
- A = (A, 0, 0)
- B = (0, B, 0)
- C = (0, 0, C)
The centroid (G) of triangle ABC is given by:
\[
G = \left(\frac{A + 0 + 0}{3}, \frac{0 + B + 0}{3}, \frac{0 + 0 + C}{3}\right) = \left(\frac{A}{3}, \frac{B}{3}, \frac{C}{3}\right)
\]
4. **Substituting A, B, C in terms of G**:
Let:
- \(x = \frac{A}{3}\)
- \(y = \frac{B}{3}\)
- \(z = \frac{C}{3}\)
Then:
\[
A = 3x, \quad B = 3y, \quad C = 3z
\]
5. **Substituting into the Distance Equation**:
Substitute \(A\), \(B\), and \(C\) into the distance equation:
\[
\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{9}
\]
This simplifies to:
\[
\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{9}
\]
Multiplying through by 9 gives:
\[
\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1
\]
6. **Final Locus Equation**:
The locus of the centroid G of triangle ABC is given by:
\[
\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1
\]
### Conclusion:
The locus of the centroid of triangle ABC is given by the equation:
\[
\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1
\]
