A sphere of constant radius $r$ passes through the origin $O$ and cuts the axes in $A,B,C.$ Find the locus of the foot of the perpendicular from $O$ to the plane $ABC$.
My solution goes like this:
We consider the intercepts by the sphere as $(A,0,0),(0,B,0),(0,0,C)$ respectively. The equation of the plane $ABC$ is $\frac xA+\frac yB+\frac zC=1$ and the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$. Substituting, $(A,0,0),(0,B,0),(0,0,C),(0,0,0)$ in the equation of the sphere, we get $A=-2u,B=-2v,C=2w$. We can write the equation of the plane as $\overrightarrow{r}(\frac {1}{A}\overrightarrow {i}+\frac {1}{B}\overrightarrow {j}+\frac {1}{C}\overrightarrow {k})=1 $ and hence, the the coordinates of the foot of perpendicular are $$\left(\frac{\frac {1}{A}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\mathbin, \frac{\frac {1}{B}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\mathbin, \frac{\frac {1}{C}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\right)\cdot$$ Now, the equation of the normal is $$\frac{x-0}{\frac 1A}=\frac{y-0}{\frac 1B}=\frac{z-0}{\frac 1C}=r_1.$$ Thus, $x=\frac {r_1}{A},y=\frac {r_1}{B},z=\frac {r_1}{C}.$ Now, since $$\left(\frac{\frac {1}{A}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2},\frac{\frac {1}{B}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2},\frac{\frac {1}{C}}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\right) =(\alpha,\beta,\gamma)(\text{say}),$$ hence $\alpha^2+\beta^2+\gamma^2=\frac{1}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}.$ Also, we know that, $u^2+v^2+w^2=r^2$ which implies $A^2+B^2+C^2=4r^2.$ Now, $(\alpha,\beta,\gamma)$ lies on the line $$\frac{x-0}{\frac 1A}=\frac{y-0}{\frac 1B}=\frac{z-0}{\frac 1C}=r_1.$$ Hence, $\alpha=\frac {r_1}{A}$, $\beta=\frac {r_1}{B}$, $\gamma=\frac {r_1}{C}$, due to which $r_1=\frac{1}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2}\cdot$ Also, $r_1=\frac{1}{(\frac1A)^2+(\frac1B)^2+(\frac1C)^2} =(\alpha^2+\beta^2+\gamma^2) =\alpha A=\beta B=\gamma C.$ So, $A=\frac{\alpha^2+\beta^2+\gamma^2}{\alpha}$, $B=\frac{\alpha^2+\beta^2+\gamma^2}{\beta}$, $C=\frac{\alpha^2+\beta^2+\gamma^2}{\gamma}\cdot$ Again, $A^2+B^2+C^2=4r^2$ thus $$\begin{eqnarray*} &&\left(\frac{\alpha^2+\beta^2+\gamma^2}{\alpha})^2\right) +\left(\frac{\alpha^2+\beta^2+\gamma^2}{\beta}\right)^2 +\left(\frac{\alpha^2+\beta^2+\gamma^2}{\gamma}\right)^2\\ &&\qquad{}={}4r^2\\ &&\qquad{}={}\left(\frac{x^2+y^2+z^2}{x}\right)^2 +\left(\frac{x^2+y^2+z^2}{y}\right)^2 +\left(\frac{x^2+y^2+z^2}{z}\right)^2\\ &&\qquad{}={}4r^2, \end{eqnarray*}$$ is the required locus.
Is the above solution correct? If not, where is it going wrong?...