If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at $$A$$, $$B$$ & $$C$$, then the locus of the centroid of $$\triangle ABC$$ is

Let the required variable plane meet the coordinate axes at the three intercepts $$A(a,0,0),\;B(0,b,0),\;C(0,0,c)$$. The equation of such a plane is most conveniently written in intercept form

$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\;.-(1)$$

The distance of a plane given by $$Px+Qy+Rz+S=0$$ from the origin $$O(0,0,0)$$ is $$\dfrac{|S|}{\sqrt{P^{2}+Q^{2}+R^{2}}}\,.$$ To use this, rewrite equation $$-(1)$$ in the general linear form:

$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}-1=0 \;\Longrightarrow\; \left(\frac{1}{a}\right)x+\left(\frac{1}{b}\right)y+\left(\frac{1}{c}\right)z-1=0\;.-(2)$$

Here $$P=\dfrac{1}{a},\;Q=\dfrac{1}{b},\;R=\dfrac{1}{c},\;S=-1$$. The given condition is that the distance of the plane from the origin equals 3 units. Applying the distance formula, we get

$$3=\frac{|\,S\,|}{\sqrt{P^{2}+Q^{2}+R^{2}}} =\frac{|\,{-1}\,|}{\sqrt{\left(\dfrac{1}{a}\right)^{2} +\left(\dfrac{1}{b}\right)^{2} +\left(\dfrac{1}{c}\right)^{2}}}\;.-(3)$$

Simplifying equation $$-(3)$$ step by step:

$$3=\frac{1}{\sqrt{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}}}$$

Taking reciprocals on both sides,

$$\sqrt{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}}=\frac{1}{3}$$

Squaring to clear the square root,

$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{9}\;.-(4)$$

Next, we determine the coordinates of the centroid of $$\triangle ABC$$. For the vertices $$A(a,0,0),\;B(0,b,0),\;C(0,0,c)$$ the centroid $$G(x,y,z)$$ is obtained by averaging the corresponding coordinates:

$$x=\frac{a+0+0}{3}=\frac{a}{3},\qquad y=\frac{0+b+0}{3}=\frac{b}{3},\qquad z=\frac{0+0+c}{3}=\frac{c}{3}\;.-(5)$$

Equation $$-(5)$$ can be inverted to express $$a,\;b,\;c$$ in terms of $$x,\;y,\;z$$:

$$a=3x,\qquad b=3y,\qquad c=3z\;.-(6)$$

Substitute these values of $$a,\;b,\;c$$ from $$-(6)$$ into the distance condition $$-(4)$$:

$$\frac{1}{(3x)^{2}}+\frac{1}{(3y)^{2}}+\frac{1}{(3z)^{2}}=\frac{1}{9}$$

Since $$(3x)^{2}=9x^{2}$$ and likewise for the other two, this becomes

$$\frac{1}{9x^{2}}+\frac{1}{9y^{2}}+\frac{1}{9z^{2}}=\frac{1}{9}\;.-(7)$$

Multiply every term in equation $$-(7)$$ by 9 to clear the common denominator:

$$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=1\;.-(8)$$

Equation $$-(8)$$ represents the required locus of the centroid $$G(x,y,z)$$ of the triangle whose intercept plane is always 3 units away from the origin.

In the language of the options provided, equation $$-(8)$$ is exactly the statement

$$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=1.$$ Hence, the correct answer is Option A.