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 $O$ 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?...

I checked your computations. I agree with implicit equation that can be written as well under the form:
$$(x^2+y^2+z^2)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=4r^2\tag{1}$$
I will take now on WLOG the value $r=\frac12$ in order that the RHS in (1) is $1$.
(1) can be given the equivalent form avoiding "divisions by zeros" :
$$(x^2+y^2+z^2)^2\left(x^2y^2+y^2z^2+z^2x^2\right)=x^2y^2z^2\tag{2}$$
which has a representation looking like orbitals in chemistry :

For those who are interested, here is the Matlab program that I have written for generating the above image :
 clear all;close all;hold on;axis equal off;
 a=0.3;I=-a:0.01:a;[x,y,z]=meshgrid(I,I,I);
 axis([-a,a,-a,a,-a,a]);
 f=((x.^2+y.^2+z.^2).^2).*((x.*y).^2+(y.*z).^2+(z.*x).^2)-(x.*y.*z).^2;
 [faces,verts,colors] = isosurface(x,y,z,f,0,abs(z));
 patch('Vertices',verts,'Faces',faces,'FaceVertexCData',colors,...
 'FaceColor','interp','EdgeColor','none');
 view([20,24]);
 plot3([0,a,0,0,0,0,0],[0,0,0,a,0,0,0],[0,0,0,0,0,0,a],'color','k')

Edit : In fact, the form of the coordinates of the projection point $$(\frac{\frac{1}{A}}{D},\frac{\frac{1}{B}}{D},\frac{\frac{1}{C}}{D})
\text{ where } D:=(\frac1A)^2+(\frac1B)^2+(\frac1C)^2)\tag{3}$$
evokes a possible connection with the geometric transform called inversion. Indeed, we can establish such a relationship. Let us consider the simplest inversion, with the origin as its center and with "power" $1$, wich means that point $M$ has image $M'$ if and only if :
$$\vec{OM'}=\frac{\vec{OM}}{\|\vec{OM}\|^2}$$
which is equivalent to
$$O,M,M' \ \text{are aligned and  } \vec{OM'}.\vec{OM'}=1$$
Let $A(a,0,0), B(0,b,0), C(0,0,c)$ (please note the slight notational changes). The image of the plane $A,B,C$ is the sphere passing through the origin $O$ and the three points
$A'(\frac{1}{a},0,0), B'(0,\frac{1}{b},0), C'(0,0,\frac{1}{c})$
This sphere passing through these 4 points is easily proven to have equation :
$$x^2+y^2+z^2-\frac{1}{a}x-\frac{1}{b}y-\frac{1}{c}z=0$$
in which we "read" the coordinates of its center
$$D(\frac{1}{2a},\frac{1}{2b},\frac{1}{2c}).$$
Now consider line $OD$ : it intersects the sphere in a point $H'$ where line $OD$ is a diameter ; therefore $OD$ is orthogonal to the sphere in $H'$ but it means that $H'$ is the image of the foot of the altitude because orthogonality is preserved by inversion. Therefore (3) is established by unicity of a point $M$ such that $OM$ is orthogonal to the sphere.
