Answer: \(\alpha=-1\).

The other eigenvalues are \(\lambda_1=6\) and \(\lambda_2=3\).

Corresponding eigenvectors are \(\begin{pmatrix}1\\2\\-2\end{pmatrix}\) for \(\lambda=-9\), \(\mathbf e_1=\begin{pmatrix}-2\\2\\1\end{pmatrix}\) for \(\lambda_1=6\), and \(\mathbf e_2=\begin{pmatrix}2\\1\\2\end{pmatrix}\) for \(\lambda_2=3\).

\(\mathbf M\mathbf x=6a\mathbf e_1+3b\mathbf e_2\), so \(p=6a\) and \(q=3b\).

Since \(-9\) is an eigenvalue, \(\det(\mathbf M+9\mathbf I)=0\).

Thus

\(\begin{vmatrix}12&-4&2\\-4&\alpha+9&6\\2&6&7\end{vmatrix}=0\).

Expanding gives \(80\alpha+80=0\), so \(\alpha=-1\).

With \(\alpha=-1\),

\(\mathbf M=\begin{pmatrix}3&-4&2\\-4&-1&6\\2&6&-2\end{pmatrix}\).

The characteristic equation is

\(\det(\mathbf M-\lambda\mathbf I)=0\),

which simplifies to

\(\lambda^3-63\lambda+162=0\).

Since \(-9\) is already a root,

\(\lambda^3-63\lambda+162=(\lambda+9)(\lambda-6)(\lambda-3)\).

Therefore the other two eigenvalues are \(\lambda_1=6\) and \(\lambda_2=3\).

Solving \((\mathbf M-\lambda\mathbf I)\mathbf e=\mathbf0\) for each eigenvalue gives the following corresponding eigenvectors:

for \(\lambda=-9\), \(\mathbf e=\begin{pmatrix}1\\2\\-2\end{pmatrix}\);

for \(\lambda_1=6\), \(\mathbf e_1=\begin{pmatrix}-2\\2\\1\end{pmatrix}\);

for \(\lambda_2=3\), \(\mathbf e_2=\begin{pmatrix}2\\1\\2\end{pmatrix}\).

Now \(\mathbf x=a\mathbf e_1+b\mathbf e_2\). Since \(\mathbf e_1\) and \(\mathbf e_2\) are eigenvectors corresponding to \(6\) and \(3\), respectively,

\(\mathbf M\mathbf e_1=6\mathbf e_1\), and \(\mathbf M\mathbf e_2=3\mathbf e_2\).

Hence

\(\mathbf M\mathbf x=\mathbf M(a\mathbf e_1+b\mathbf e_2)=a\mathbf M\mathbf e_1+b\mathbf M\mathbf e_2=6a\mathbf e_1+3b\mathbf e_2\).

Therefore \(p=6a\) and \(q=3b\).