A variable plane is at a constant distance p from the origin O and meets the set of rectangular axes $OX_i (i = 1, 2, 3)$ at points $A_i (i=1, 2, 3)$ respectively. If planes are drawn through $A_1, A_2, A_3)<$ which are parallel to the coordinate planes, then the locus of their point intersection, is

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

Let the equation of the plane be $\frac{x}{a} +\frac{y}{b}+\frac{z}{c}= 1.$ It meets the coordinates axes at $A_1(a, 0, 0), A_2(0, b, 0)$ and $A_3(0, 0, c)$. The equations of planes passing through coordinate planes are $x = a, y = b$ and $z=c$. Let $(\alpha, \beta , \gamma )$ be the point of intersection of these planes. Then, $\alpha =a, \beta= b, \gamma = c.$ The plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 1$ is at a distance p from the origin.

$∴p = \begin{vmatrix} \frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\end{vmatrix}$

$⇒\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$        $[∵ a = \alpha, b = \beta \gamma = c]$

$⇒\frac{1}{p^2}=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$

Hence, the locus of $(\alpha , \beta , \gamma )$ is $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}.$