A variable plane makes intercepts on the co – ordinate axes such that the sum of their squares is constant and equal to $k^2.$ Then the locus of the foot of the perpendicular from the origin to the plane is,

Let us take a variables plane $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1.$ It intersect the co – ordinate axes at the points A, B and C, $A=(p,o,o),B=(o,q,o),C=(o,o,r).$ Given that $p^2+q^2+r^2=k^2.$ Now we have to find the locus of foot of normal from the origin to the plane. Let it  be N. Equation of ON $=\frac{x}{1/p}=\frac{y}{1/q}=\frac{z}{1/r}=\lambda$
$(\frac{\lambda }{p},\frac{\lambda }{q},\frac{\lambda }{r})$ lies on the plane

$\therefore \frac{\lambda }{p^2}+\frac{\lambda }{q^2}+\frac{\lambda }{r^2}=1\Rightarrow \lambda =\frac{1}{{{\displaystyle \sum p}}^{-2}}$

Foot of the normal N $=(\frac{p^{-1}}{{\displaystyle \sum p^{-2}}},\frac{q^{-1}}{{\displaystyle \sum p^{-2}}},\frac{r^{-1}}{{\displaystyle \sum p^{-2}}})$

$x=\frac{p^{-1}}{{\displaystyle \sum p^{-2}}},y=\frac{q^{-1}}{{\displaystyle \sum p^{-2}}},z=\frac{r^{-1}}{{\displaystyle \sum p^{-2}}}$

$x^2+y^2+z^2=\frac{p^{-2}+q^{-2}+r^{-2}}{({\displaystyle \sum p^{-2}{)}^2}}=\frac{1}{{\displaystyle \sum p^{-2}}} \dots \dots \dots \dots ..\left(ii\right)$

$x^{-2}+y^{-2}+z^{-2}=(p^2+q^2+r^2)(\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}) \dots \dots .\left(iii\right)$

From equation (ii) and equation (iii) we get  ${(x^2+y^2+z^2)}^2(x^{-2}+y^{-2}+z^{-2})=p^2+q^2+r^2=k^2$