Step 1
Given Matrix \[A = \left[ {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right]\]Eigenvector:\[{\bf{v}} = \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right]\]We have to find the corresponding eigenvalue for the above eigenvector.
An eigenvector of an $n \times n$ matrix A is a nonzero vector x such that $A{\bf{x}} = \lambda {\bf{x}}$ for some scalar $\lambda $. A scalar $\lambda $ is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = \lambda {\bf{x}}$; such an $\bf{x}$ is called an eigenvector corresponding to $\lambda $.



Step 2: The equation form
Write the equation to find eigenvalues.\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\\left[ {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{3 \times 1 + 6 \times \left( { - 2} \right) + 7 \times 1}\\{3 \times 1 + 3 \times \left( { - 2} \right) + 7 \times 1}\\{5 \times 1 + 6 \times \left( { - 2} \right) + 5 \times 1}\end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{3 - 12 + 7}\\{3 - 6 + 7}\\{5 - 12 + 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}\lambda \\{ - 2\lambda }\\\lambda \end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{ - 2}\\4\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}\lambda \\{ - 2\lambda }\\\lambda \end{array}} \right]\end{array}\]

ANSWER
As from the above matrix form of equation that the only possible solution is when $\lambda = -2$.

$\lambda = -2$ is an eigenvalue for the matrix and corresponding eigenvector.