= Solution
Two real symmetric <bilinear forms> are congruent if an invertible <linear map> takes one to the other; in matrices, $B=S^{\mathsf T}AS$ for some invertible $S$. The <spectral theorem for real symmetric matrices> gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form to
$$
\operatorname{diag}(I_p,-I_q,0_r),\qquad p+q+r=n.
$$
<Sylvester's law of inertia> asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.
The <rank of a quadratic form> is $p+q$, and its <signature of a quadratic form> is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\geq0$ with $p+q\leq n$ gives
$$
\boxed{\text{number of classes}=\sum_{p=0}^n(n-p+1)=\frac{(n+1)(n+2)}2.}
$$
For $n\geq1$, \b[exactly two classes have every level set bounded]: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.
For the displayed three-dimensional form,
$$
A=\begin{pmatrix}3&2&1\\2&6&4\\1&4&5\end{pmatrix}.
$$
Its leading principal minors are $3,14,32$, all positive. <Sylvester's criterion> makes it positive definite, so \b[the surface is a bounded ellipsoid]. Directly, its quadratic expression is
$$
3\left(x+\frac23y+\frac13z\right)^2
+\frac{14}{3}\left(y+\frac57z\right)^2+\frac{16}{7}z^2,
$$
which also proves positivity and boundedness of the unit level.
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