The correct answer is $\boxed{\text{A}}$.
The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a matrix $A$ is given by $$p(x) = |xI – A|.$$
In this case, we have $$p(x) = |xI – A| = \begin{vmatrix} x – 1 & 2 & 3 \\ 0 & x – 4 & 7 \\ 0 & 0 & x – 3 \end{vmatrix} = (x – 1)(x – 4)(x – 3).$$
Therefore, the eigenvalues of the matrix are $1$, $4$, and $3$.
Option A is the only option that contains all three of these eigenvalues.