If the sphere intersects the three axes at the points A, B and C, it means that the sphere passes through these three points:
$(A,0,0)$; $(0,B,0)$; $(0,0,C)$
i.e., the sphere intersects x-axis at $(A,0,0)$ and y-axis at $(0,B,0)$ and z-axis at $(0,0,C)$
Now, coordinates of the centroid of triangle formed by these three co-ordinates will be:
$(\frac{A+0+0}{3},\frac{0+B+0}{3},\frac{0+0+C}{3})$ = $(\frac{A}{3},\frac{B}{3},\frac{C}{3})$
To find distance of this centroid from origin $(0,0,0)$ we simply use the two point distance formula in 3d:
Distance between $(\frac{A}{3},\frac{B}{3},\frac{C}{3})$ and $(0,0,0)$ =
$\sqrt{(\frac{A}{3}-0)^2 + (\frac{B}{3}-0)^2 + (\frac{C}{3}-0)^2}$ = $\frac{\sqrt{A^2 + B^2 + C^2}}{3}$
In our case, the radius of sphere is given as $r$.
Let's assume the centre of sphere to be at $(x_0,y_0,z_0)$. Then the equation of the sphere can be written as:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = r^2$
Since it's also given that the sphere passes through the origin $(0,0,0)$, this point must satisfy the above equation of sphere.
Or, $(0-x_0)^2+(0-y_0)^2+(0-z_0)^2 = r^2$
Or, $x_0^2 + y_0^2 + z_0^2 = r^2$ --- $(1)$
Now, the sphere also passes through the point $(A,0,0)$. Substituting this point in the equation of sphere, we have:
$(A-x_0)^2+(0-y_0)^2+(0-z_0)^2 = r^2$
Or, $(A-x_0)^2+y_0^2+z_0^2 = r^2$
From $(1)$, let's also substitute $r^2$ with $x_0^2 + y_0^2 + z_0^2$
$(A-x_0)^2+y_0^2+z_0^2 = x_0^2 + y_0^2 + z_0^2$
Or, $(A-x_0)^2 = x_0^2$
Or, $A-x_0 = x_0$
Thus we have $A = 2x_0$.
Similarly by substituting $(0,B,0)$ and $(0,0,C)$ in the equation of sphere we get $B = 2y_0$ and $C= 2z_0$.
Hence, the three points A, B and C are $(2x_0,0,0)$, $(0,2y_0,0)$ and $(0,0,2z_0)$ respectively.
Thus, by using the formula we arrived at in section 1, we can say the distance of centroid of triangle ABC from origin =
$\frac{\sqrt{A^2 + B^2 + C^2}}{3}$ = $\frac{\sqrt{(2x_0)^2 + (2y_0)^2 + (2z_0)^2}}{3}$ = $\frac{2\sqrt{x_0^2 + y_0^2 + z_0^2}}{3}$
But from equation $(1)$ we know that $x_0^2 + y_0^2 + z_0^2 = r^2$.
Substituting this we have the required distance as $\frac{2\sqrt{r^2}}{3} = \frac{2r}{3}$
Best Answer
The equation of a variable plane in intercept form is $$\frac xa+\frac yb+ \frac zc=1$$
Here, $a,b$ and $c$ are intercept on $x,y$ and $z$ axes.
Given, it's distance from origin is constant $$\frac{\Big|\Big(\frac xa+\frac yb+ \frac zc-1\Big)_{x=y=z=0}\Big|}{\sqrt{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}}}=p(\text{constant})$$
And centroid of triangle $$(x,y,z)=\left(\frac a3,\frac b3,\frac c3\right)$$
Therefore you get $$\frac{1}{p^2}=\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}$$
Also, $a=3x,b=3y$ and $c=3z$
$$\implies \frac{1}{p^2}=\frac {1}{9x^2}+\frac {1}{9y^2}+\frac {1}{9z^2}$$
Hence the desired locus is : $$\color{blue}{\frac {1}{x^2}+\frac {1}{y^2}+\frac {1}{z^2}=\frac{9}{p^2}}$$