By definition, if $\vec{v}$ is an eigenvector of matrix $A$ corresponding to an eigenvalue $\lambda$, then they must satisfy the relation:
\[ A\vec{v} = \lambda \vec{v} \]
Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix}$ and the eigenvector $\vec{v} = \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$:
Calculate the product $A\vec{v}$:
\[ A\vec{v} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} \]
\[ A\vec{v} = \begin{bmatrix} (1 \times 2) + (2 \times 2) + (3 \times 2) \\ (1 \times 2) + (2 \times 2) + (3 \times 2) \\ (1 \times 2) + (2 \times 2) + (3 \times 2) \end{bmatrix} \]
\[ A\vec{v} = \begin{bmatrix} 2 + 4 + 6 \\ 2 + 4 + 6 \\ 2 + 4 + 6 \end{bmatrix} = \begin{bmatrix} 12 \\ 12 \\ 12 \end{bmatrix} \]
Now, express the result in terms of $\lambda \vec{v}$:
\[ \begin{bmatrix} 12 \\ 12 \\ 12 \end{bmatrix} = \lambda \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix} \]
\[ 12 = 2\lambda \implies \lambda = 6 \]
