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.

What you need is a second (linearly independent) eigenvector associated with $\lambda = 1$.  That is, you need to find another solution $x$ to 
$$
(A - I)x = 0
$$
What you were looking for was a solution to 
$$
(A - I)x = v_1
$$
which would be a generalized eigenvector.  Generalized eigenvectors are only useful when the matrix in question fails to be diagonalizable, which is not the case here (since we are meant to diagonalize it).
In fact, in order to have $U^TU = I$, you must select the eigenvectors for $\lambda = 1$ to be both length $1$ and mutually orthogonal.
