= Solution
The <quadratic form> is $\mathbf x^TQ\mathbf x$ for the real <symmetric matrix>
$$
Q=\begin{pmatrix}5&0&3\\0&-3&0\\3&0&-3\end{pmatrix}.
$$
The $y$ axis has <eigenvalue> $-3$. On the $xz$ plane, the <characteristic polynomial> is
$$
\det\begin{pmatrix}5-\ell&3\\3&-3-\ell\end{pmatrix}=(5-\ell)(-3-\ell)-9=\ell^2-2\ell-24=(\ell-6)(\ell+4).
$$
Thus the <orthogonal diagonalization> from part (a) yields
$$
\boxed{\lambda=6,\qquad\mu=-3,\qquad\nu=-4,\qquad6u^2-3v^2-4w^2=\alpha.}
$$
For later use, one possible <orthonormal basis> is
$$
e_u=\frac{(3,0,1)}{\sqrt{10}},\qquad e_v=(0,1,0),\qquad e_w=\frac{(1,0,-3)}{\sqrt{10}}.
$$
Then $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the <Euclidean norm> is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.
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