Step 1: Given matrix:
\[
A =
\begin{bmatrix}
2 & 1 & 1
1 & 4 & 1
1 & 1 & 2
\end{bmatrix}
\]
The characteristic equation is obtained by solving:
\[
\det(A - \lambda I) = 0
\]
Step 2: Expanding the determinant:
\[
\begin{vmatrix}
2 - \lambda & 1 & 1
1 & 4 - \lambda & 1
1 & 1 & 2 - \lambda
\end{vmatrix}
= 0
\]
Expanding along the first row:
\[
(2 - \lambda)
\begin{vmatrix}
4 - \lambda & 1
1 & 2 - \lambda
\end{vmatrix}
- 1
\begin{vmatrix}
1 & 1
1 & 2 - \lambda
\end{vmatrix}
+ 1
\begin{vmatrix}
1 & 4 - \lambda
1 & 1
\end{vmatrix}
\]
Step 3: Solving the determinant:
\[
(2 - \lambda) [(4 - \lambda)(2 - \lambda) - 1] - 1(2 - \lambda - 1) + 1(1 - 4 + \lambda)
\]
Solving the resulting cubic equation:
\[
(\lambda - 1)(\lambda - 2)(\lambda - 5) = 0
\]
Conclusion: The eigenvalues are \( 1, 2, 5 \), which corresponds to option (A).