Chapter 7 - Eigenvalues and Eigenvectors - 7.3 Symmetric Matrices and Orthogonal Diagonalization - 7.3 Exercises - Page 370: 18

See below.

The characteristic equation of the matrix is $det(xI_2-A)= det\left(\begin{bmatrix} x-2& 1&1 \\ 1& x-2&1\\ 1&1&x\\ \end{bmatrix} \right)=0$ Hence $x^3-6x^2+9x=0\\x(x-3)^2=0$ Thus the eigenvalues are $x=0$ with multiplicity $2$ and $x=3$ with multiplicity 2. A is symmetric; thus by Theorem 7.7, the corresponding $x=0$'s eigenspace will have a dimension of $1$, the corresponding $x=3$'s eigenspace will have a dimension of $2$.