The correct answer is $\boxed{\text{(C)}}$.
To find the eigenvalues of a matrix, we can use the following formula:
$$\lambda = \det(A – \lambda I)$$
where $A$ is the matrix and $\lambda$ is the eigenvalue.
In this case, we have:
$$\begin{align}
\det(A – \lambda I) &= \det \left[ {\begin{array}{{20}{c}} 2&{ – 2}&3 \ { – 2}&{ – 1}&6 \ 1&2&0 \end{array}} – \lambda \begin{array}{{20}{c}} 1&0&0 \ 0&1&0 \ 0&0&1 \end{array} \right] \
&= \det \left[ {\begin{array}{{20}{c}} 2-\lambda &{ – 2}&3 \ { – 2}&{ – 1}-\lambda &6 \ 1&2&-\lambda \end{array}} \right] \
&= -\lambda^3 + 4\lambda^2 – 23\lambda + 120 \
&= (\lambda – 3)(\lambda^2 – 4\lambda + 40)
\end{align*}$$
Since $\lambda = 3$ is a root of the quadratic polynomial $\lambda^2 – 4\lambda + 40$, then the other two eigenvalues must be the roots of the quadratic polynomial $(\lambda – 3)(\lambda^2 – 4\lambda + 40) = 0$.
Solving for the roots of this quadratic polynomial, we find that the other two eigenvalues are $\boxed{2, 5}$.