= Solution
{c}
Write $N=(n_1,n_2,n_3)$. Restriction to the <plane> $z=0$ gives
$$
M_{11}x^2+(M_{12}+M_{21})xy+M_{22}y^2+n_1x+n_2y+k=0.
$$
This is an equation of <degree> at most two in two <Cartesian coordinates>, the general algebraic form of a <conic>, with possible <degenerate conics> and an <empty set> of real points. No <determinant> hypothesis is needed for this substitution.
For an arbitrary <plane>, choose a point $\mathbf x_0$ and an <orthonormal basis> $\mathbf e,\mathbf f$ of its direction <linear subspace>. Put $B=(\mathbf e\ \mathbf f)$ and $\mathbf x=\mathbf x_0+B\boldsymbol\xi$, where $\boldsymbol\xi=(s,t)^T$. With $S=(M+M^T)/2$, the <symmetric part of a matrix>, the restricted equation is
$$
\boxed{\boldsymbol\xi^TB^TSB\boldsymbol\xi+(2\mathbf x_0^TSB+NB)\boldsymbol\xi+\mathbf x_0^TS\mathbf x_0+N\mathbf x_0+k=0.}
$$
This proves the plane-section result whenever “<conic>” allows the appropriate degeneracies. A <rigid motion> taking the chosen <plane> to $z=0$ gives the same argument.
\b[The source's assumption $\det M\ne0$ does not guarantee a nondegenerate <conic>.] Even with $M=I$ and $f=x^2+y^2+z^2-1$, the section $z=1$ is a single <point>, and $z=2$ has no real points. With the nonsingular <symmetric matrix> giving $f=2xz+y^2$, the section $z=0$ is the repeated <straight line> $y=0$.
There is a further literal omission: the source does not assume $M$ is symmetric. For
$$
M=\begin{pmatrix}0&1&1\\-1&0&1\\1&-1&0\end{pmatrix},\qquad N=0,\qquad k=0,
$$
one has $\det M=2$ but $f=2xz$. Thus the section $z=0$ is the entire <plane>, rather than a <conic>. Only $S$ controls the <quadratic form>; invertibility of an arbitrary $M$ cannot remove this exception. If the intended hypothesis is that $M$ is symmetric and nonsingular, the quadratic restriction to a <plane> cannot vanish identically: otherwise $B^TMB=0$, so $M$ would map a two-dimensional <linear subspace> injectively into its one-dimensional <orthogonal complement>, which is impossible. Degenerate and empty sections still remain possible.
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