If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A,B anc C, then the locus of the centroid of `Delta ABC` is :

Options

A. `(1)/(x ^(2)) + (1)/(y ^(2)) + (1)/(z ^(2)) =1`
B. `(1)/(x ^(2)) + (1)/(y ^(2)) + (1)/(z ^(2)) =3`
C. `(1)/(x ^(2)) + (1)/(y ^(2)) + (1)/(z ^(2)) =9`
D. `(1)/(x ^(2)) + (1)/(y ^(2)) + (1)/(z ^(2)) =1/9`

Answer

To solve the problem step by step, we need to find the locus of the centroid of triangle ABC formed by the intersections of a variable plane with the coordinate axes, given that the plane is at a distance of 3 units from the origin.

### Step 1: Set up the equation of the plane
Let the equation of the variable plane be given in the intercept form:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \( A(a, 0, 0) \), \( B(0, b, 0) \), and \( C(0, 0, c) \) are the points where the plane intersects the x, y, and z axes respectively.

### Step 2: Calculate the distance from the origin to the plane
The distance \( d \) from the origin to the plane can be calculated using the formula:
\[ d = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \]
According to the problem, this distance is given as 3 units. Thus, we have:
\[ \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3 \]

### Step 3: Rearranging the equation
Squaring both sides gives:
\[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9} \]
This can be rewritten as:
\[ 9\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right) = 1 \]

### Step 4: Find the coordinates of the centroid
The coordinates of the centroid \( G \) of triangle \( ABC \) are given by:
\[ G\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \]

### Step 5: Express \( a, b, c \) in terms of \( G \)
Let:
\[ x = \frac{a}{3}, \quad y = \frac{b}{3}, \quad z = \frac{c}{3} \]
Thus, we can express \( a, b, c \) as:
\[ a = 3x, \quad b = 3y, \quad c = 3z \]

### Step 6: Substitute into the distance equation
Substituting \( a, b, c \) into the distance equation gives:
\[ 9\left(\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2}\right) = 1 \]
This simplifies to:
\[ \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1 \]

### Conclusion
The locus of the centroid \( G \) is given by the equation:
\[ \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1 \]
This represents a surface in three-dimensional space.