I've attached a text that explains the Dirac braket notation and the matrix elements.

You can find the eigenvalues in the straightforward way be setting Det[O - lambda I] equal to zero and solving for lambda. But it can be done in a simpler way by noting that O maps the linear space spanned by |1> and |2> onto itself and maps the linear space spanned by |3> onto itself (see part c) of the problem for the way |1>, |2>, and |3> are defined).
We have

O|1> = |1> - i |2>

O|2> = i |1> + |2>

O|3> = 3 |3>

So, the vector |3> is an eigenvector with eigenvalue 3. We only need to diagonalize ...

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