Can't calculate an eigenvector because the system has no solutions. Find all eigenvalues of the matrix $A=\begin{pmatrix}
2&2&-2\\
2&5&-4\\
-2&-4&5\\
                             \end{pmatrix}$
and find matrix $U$ such that $U^TU=I$ and $U^TAU$ is diagonal.
I calculated and checked that $\lambda_1=10,\lambda_{2,3}=1$. Then eigenvector if $\lambda_1=10$ is  $(1,2,-2)^T$; if $\lambda_{2}=1$ eigenvector is $(0,1,1)^T$. How to find the third eigenvector (if I try to calculate the third eigenvector in this way $(A-I)x=(0,1,1)^T$, the system has no solutions.

The two eigenvectors you already found are orthogonal. Take their cross product and you'll have the third.
