Your given matrix is $\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{bmatrix}$. The eigenvalue equation is $\left|\begin{array}{ccc} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 1 & -3 & 3- \lambda \end{array}\right|= -\lambda \left|\begin{array}{cc} -\lambda & 1 \\ -2 & -3- \lambda \end{array}\right|- \left|\begin{array}{cc}0 & 1 \\ 1 & 3- \lambda\end{array}\right|=-(\lambda- 1)^3= 0$. Yes, $\lambda= 1$ is an eigenvalue with "algebraic multiplicity" 3.

An "eigenvector", v, corresponding to eigenvalue $\lambda$, of a linear transformation, A, satisfies $Av= \lambda v$. Here, that means that an eigenvector, $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$, corresponding to eigenvalue 1 satisfies $\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix} y \\ z \\ x- 3y+ 3z\end{bmatrix}= \begin{bmatrix x \\ y \\ z\end{bmatrix}$. So we have the three equations y= x, z= y, and x- 3y+ 3z= z. From the first three equations, x= y= z so the third equation can be written z- 3z+ 3z= z which is true for all z. Any eigenvector can be written in the form $\begin{bmatrix} z \\ z \\ z \end{bmatrix}= z\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}$.

Since any eigenvector is a multiple of that single vector, the "geometric multiplicity" of eigenvalue 1 is only 1.

Personally, I would not have written this problem in terms of matrices at all! The original differential equation is y'''- 3y''+ 3y'- y= 0 which has "characteristic equation" $r^3- 3r^2+ 3r- 1= (r- 1)^3= 0$ (Yes, that is the same as the eigenvalue equation). That means that the general solution to the differential equation is $y(x)= Ae^x+ Bxe^x+ Cx^2e^x= (Cx^2+ Bx+ A)e^x$.