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Split the mixed quadratic terms evenly across a real symmetric matrix:Its characteristic polynomial isso $c=2$. Choose the orthonormal eigenvectorsThe orthogonal transformation $\mathbf{r}=P\mathbf{R}$ givesIndeed $P^{T}P=I$ and $P^{T}AP=\mathrm{diag}(-1,-5,4)$; preserving $Q_2$ is precisely the length-preserving property of an orthogonal matrix.

The requested plane has normal $(2,1,-2)$. An orthonormal basis within it isWrite a point of that plane as $\mathbf{r}=u{e}_1+w{e}_2$. Restricting the quadratic form gives $Q_1=-u^2-(13/9)w^2$, so its section with $Q_1=-5$ isIt is an ellipse centred at the origin, with semi-axes $\sqrt{5}$ along $e_1$ and $\sqrt{45/13}$ along $e_2$. Normalizing the in-plane basis is essential: coefficients in nonorthonormal coordinates would not directly give physical lengths.

The complete surface is the one-sheet hyperboloidIt is connected, extends indefinitely in both directions along its axis $u_3$, and has an elliptical waist at $Z=0$ with semi-axes $\sqrt{5}$ and $1$. At larger $|Z|$ its transverse sections widen; its asymptotic cone is $4Z^2=X^2+5Y^2$. The specified plane is oblique to that axis, rather than the waist plane.

Interpret the norms as Euclidean norms and set $B=A^{\mathsf{T}}A$. This is a real symmetric matrix with ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}=\Vert A\mathbf{x}{\Vert }^2\ge 0$, hence a positive semidefinite matrix. Squaring the nonnegative objective does not change its argmin. Let ${\lambda }_{\min }$ be the smallest eigenvalue of $B$ and $E_{\min }$ its eigenspace.

First minimize on the unit sphere. The Lagrange multiplier function $L(\mathbf{x},\lambda )={\mathbf{x}}^{\mathsf{T}}B\mathbf{x}-\lambda ({\mathbf{x}}^{\mathsf{T}}\mathbf{x}-1)$ has stationarity condition $B\mathbf{x}=\lambda \mathbf{x}$. Multiplication by ${\mathbf{x}}^{\mathsf{T}}$ shows that the squared objective at such a point is $\lambda$. The spectral theorem for real symmetric matrices also gives, for every unit vector, ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}\ge {\lambda }_{\min }$, with equality exactly in $E_{\min }$. This proves global minimality, rather than merely finding stationary candidates.

If ${\lambda }_{\min }>0$, every point with $\Vert \mathbf{x}\Vert >1$ has a strictly smaller objective after rescaling to unit length. ThereforeIf ${\lambda }_{\min }=0$, the kernel of a linear map satisfies $\ker B=\ker A$, since $B\mathbf{x}=0$ implies $\Vert A\mathbf{x}{\Vert }^2=0$, and the converse is immediate. The minimum is then zero, attained precisely atIn this case minimizers need not have unit length. This distinction is essential for a rank-deficient $A$, including any real $m\times n$ matrix with $m<n$. The result is the minimum image norm outside a Euclidean unit ball, governed by the least singular value.