All I know is because this matrix is an upper triangular matrix, the eigenvalues are $2$ and $-5$. If I'm not mistaken, when the eigenvalue is $2$, rank of $A-2I$ must be $1$ in order to get $2$ eigenvectors but I don't understand why. Also, I don't know how to solve this question.
Thanks!!
In order for your matrix to be diagonalizable, it's enough to find a basis consisting of eigenvectors. Since the eigenvalues are $2$ and $-5$ with multiplicities $2$ and $1$ respectively, and the dimension of the space must be equal to the sum of the dimensions of distinct eigen spaces, so $$3=dim(E_{2})+dim(E_{-5})$$ where $dim(E_{\lambda})$ denotes the dimension of the eigen spavce corresponding to $\lambda.$ Since $dim(E_{-5})=1$ you must have $dim(E_{2})=2$ for your matrix to be diagonalizable.I think this answers your question "why rank$(A-2I)$ must be $1.$"