A plane a constant distance p from the origin meets the coordinate axes in A, B, C. Locus of the centroid of the triangle ABC is

Options

A. `x^(2)+y^(2)+z^(2)=p^(2)`
B. `x^(-2)+y^(-2)+z^(-2)=9p^(-2)`
C. `x^(-1)+y^(-1)+z^(-1)=3p^(-1)`
D. none of these

Answer

To find the locus of the centroid of triangle ABC formed by the intersection of a plane at a constant distance \( p \) from the origin with the coordinate axes, we can follow these steps:

### Step 1: Equation of the Plane
The equation of a plane in intercept form that meets the coordinate axes at points A, B, and C can be expressed as:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \( a, b, c \) are the x-intercept, y-intercept, and z-intercept of the plane, respectively.

### Step 2: Distance from the Origin
The distance \( d \) from the origin to the plane is given by the formula:
\[ d = \frac{|1|}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2}} = p \]
Squaring both sides, we have:
\[ p^2 = \frac{1}{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2} \]
This simplifies to:
\[ \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \]

### Step 3: Centroid of Triangle ABC
The coordinates of the centroid \( G \) of triangle ABC can be calculated using the formula:
\[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right) \]
For points A, B, and C, we have:
- \( A(a, 0, 0) \)
- \( B(0, b, 0) \)
- \( C(0, 0, c) \)

Thus, the coordinates of the centroid are:
\[ G = \left(\frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \]

### Step 4: Expressing \( a, b, c \) in terms of \( G \)
Let the coordinates of the centroid \( G \) be \( (x, y, z) \). Then:
\[ x = \frac{a}{3} \implies a = 3x \]
\[ y = \frac{b}{3} \implies b = 3y \]
\[ z = \frac{c}{3} \implies c = 3z \]

### Step 5: Substituting Back into the Distance Equation
Substituting \( a, b, c \) into the distance equation:
\[ \frac{1}{p^2} = \frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} \]
This simplifies to:
\[ \frac{1}{p^2} = \frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} \]
Multiplying through by \( 9 \):
\[ \frac{9}{p^2} = \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \]

### Step 6: Final Locus Equation
Rearranging gives us the locus of the centroid:
\[ \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{p^2} \]