Given a three dimensional Hilbert space with the three basis vectors $|1\rangle, |2\rangle, |3\rangle,$ and two state vectors, $|\psi_1 \rangle = a|1\rangle -b |2\rangle +c|3\rangle, |\psi_2 \rangle = b|1\rangle + a |2\rangle,$ with $a,b,c \in \mathbb{C},$ I need to find if the operator $A = |\psi_1' \rangle \langle \psi_1'| + |\psi_2' \rangle \langle \psi_2'|$ is a hermitian operator, whereby $|\psi_1' \rangle, |\psi_2' \rangle,$ are the corresponding normed vectors. In case the scalars $a,b,c$ were reals, the matrix of $A$ would be symmetric as sum of two symmetric matrices and thus hermitian, otherwise I do not see if $A$ can be hermitian. Can somebody provide some insight  ? Many thanks.

Given a three dimensional Hilbert space with the three basis vectors $|1\rangle, |2\rangle, |3\rangle,$ and two state vectors, $|\psi_1 \rangle = a|1\rangle -b |2\rangle +c|3\rangle, |\psi_2 \rangle = b|1\rangle + a |2\rangle,$ with $a,b,c \in \mathbb{C},$ I need to find if the operator $A = |\psi_1' \rangle \langle \psi_1'| + |\psi_2' \rangle \langle \psi_2'|$ is a hermitian operator, whereby $|\psi_1' \rangle, |\psi_2' \rangle,$ are the corresponding normed vectors. In case the scalars $a,b,c$ were reals, the matrix of $A$ would be symmetric as sum of two symmetric matrices and thus hermitian, otherwise I do not see if $A$ can be hermitian. Can somebody provide some insight?