A variable plane which remains at a constant distance $3 p$ from the origin cuts the coordinate axes at $A, B, C$ . Show that the locus of the centroid of triangle $A B C$ is $\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{9}{p^{2}}$ .

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Solution

The equation of a variable plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

it's distance from origin is constant

$\frac{| (\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1) |}{\sqrt{(\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}})}} = c o n s t a n t (p^{2})$

And centroid of triangle $(x, y, z) = (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$

Therefore we get

$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$

Also, $a = 3 x, b = 3 y$ and $c = 3 z$

$⟶ \frac{1}{p^{2}} = \frac{1}{{9 x}^{2}} + \frac{1}{9 y^{2}} + \frac{1}{9 z^{2}}$

$\frac{9}{p^{2}} = \frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}}$

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