The correct answer is $\boxed{\text{A) }1, 1, 1}$.
To find the eigenvalues of a matrix, we can use the following formula:
$$\lambda = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $a$, $b$, and $c$ are the coefficients of the characteristic polynomial of the matrix.
The characteristic polynomial of the matrix $A$ is given by:
$$p(x) = |A – xI| = \det \left[ {\begin{array}{*{20}{c}} 3&{ – 1}&{ – 1} \ { – 1}&3&{ – 1} \ { – 1}&{ – 1}&3 \end{array}} \right] = x^3 – 9x^2 + 6x – 3$$
To find the eigenvalues of $A$, we need to solve the equation $p(x) = 0$.
Solving $p(x) = 0$, we get the following eigenvalues:
$$\lambda = 1, 1, 1$$
Therefore, the eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$.
Here is a brief explanation of each option:
- Option A: The eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$. This can be verified by substituting $x = 1$ into the characteristic polynomial $p(x)$.
- Option B: The eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$. This can be verified by substituting $x = 2$ into the characteristic polynomial $p(x)$.
- Option C: The eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$. This can be verified by substituting $x = 4$ into the characteristic polynomial $p(x)$.