Three points $P,Q,R$ taken on ellipsoid $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}+ \dfrac{z^2}{c^2} = 1$ so that lines joining $P,Q,R$ to origin are mutually perpendicular.
Prove that plane $PQR$ touches a fixed sphere.
Attempt : I assumed the points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ as $P,Q,R$ and so the Direction ratios of $OP,OQ,OR$ would be same as their points also the relations $x_1x_2+y_1y_2+z_1z_2=0$ etc. holds as they are mutually perpendicular.
Now I wrote the equation of plane through these three points in vector form but can not establish the relation asked.
Is the approach wrong or needs to be continued in more directed way. Any help appreciated.
We can write the equation of the ellipsoid in matrix form: $x^TMx=1$, where $$ M=\pmatrix{1/a^2 & 0 & 0\\0 & 1/b^2 & 0\\0 & 0 &1/c^2}. $$
If we rotate the cartesian frame so that $P$, $Q$, $R$ are the intersections of the ellipsoid with the axes, the equation becomes $x^TM'x=1$, where $M'$ is the transformed of $M$ under the rotation. We can then easily find the coordinates of $P$, $Q$, $R$: $$ P=((M'_{11})^{-1/2},0,0);\quad Q=(0,(M'_{22})^{-1/2},0);\quad R=(0,0,(M'_{33})^{-1/2}). $$
The distance $r$ of plane $PQR$ from the origin $O$ is the height of pyramid $OPQR$ and can be readily computed: $$ \tag{1} \DeclareMathOperator{\Tr}{Tr} r=\left(M'_{11}+M'_{22}+M'_{33}\right)^{-1/2}= \Tr(M')^{-1/2}. $$
But under a rotation the trace of a matrix doesn't change, hence we find:
$$ \tag{2} r=\Tr(M)^{-1/2}= \left({1\over a^2}+{1\over b^2}+{1\over c^2}\right)^{-1/2}. $$
This proves that plane $PQR$ has always the same distance $r$ from the origin, as requested.