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By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.

For a real symmetric matrix $A$, find its real eigenvalues from $\det (A-\lambda I)=0$, then solve $(A-\lambda I)v=0$ for the corresponding eigenspaces. Distinct eigenspaces are orthogonal. Within each repeated eigenspace, apply the Gram-Schmidt process and normalize to obtain an orthonormal basis of eigenvectors.

Put these vectors in the columns of $O$. The spectral theorem for real symmetric matrices guarantees a full basis, so $O^{T}O=I$ andThis is the orthogonal diagonalization of $A$ and the associated orthogonal change of coordinates.

The orthogonal change of coordinates preserves the sphere equation, giving $u^2+v^2+w^2=4$. Subtract it from the spheroid equation in part (b):Thus the intersection consists of the two circlesTheir common radius is $\boxed{\sqrt{5/8}=\sqrt{10}/4}$. Transforming their centers back to the original coordinates givesTheir planes have equations $x_1+x_2=\pm 3\sqrt{3}/2$, each with normal vector $(1,1,0)$. The difference ${\mathbf{c}}_{+}-{\mathbf{c}}_{-}$ is parallel to this normal, proving that the line joining the centers is perpendicular to both planes. This is the intersection of a prolate spheroid and a concentric sphere in the regime between the short and long semiaxes.

The orthogonal change of coordinates preserves distance to the origin. On the required quadric surface,Thus the squared distance is at least $1/4$. Equality holds precisely when $u=v=0$, and the constraint then gives $w=\pm 1/2$. Using $e_w=(1,0,-3)/\sqrt{10}$ from part (b), the nearest points in the original coordinates areSubstitution checks the constraint directly: $5x^2-3z^2+6xz=(5-27-18)/40=-1$. The inequality proves these are global nearest points, rather than merely stationary candidates.