A= $ \left[ \begin{array}{ccc}3i & -1+i & 3-2i\\1+i & -i & 1+2i \\-3-2i & -1+2i & 0\end{array} \right] \\ \; \\ \; \\ \therefore A' \;=\; \left[ \begin{array}{ccc} 3i & 1+i & -3-2i \\ -1+i & -i & -1+2i \\ 3-2i & 1+2i & 0\end{array} \right] \\ \; \\ \; \\ \therefore A^{\theta} \;=\; (\bar{A'}) \;=\; \left[ \begin{array}{ccc} -3i & 1-i & -3+2i \\ -1-i & i & -1-2i \\ 3+2i & 1-2i & 0 \end{array} \right] \\ \; \\ \; \\ \; \\ Let, \; P\;=\; \dfrac{1}{2}(A+A^{\theta}) \;= \; \dfrac{1}{2} \left[ \begin{array}{ccc}3i & -1+i & 3-2i\\1+i & -i & 1+2i \\-3-2i & -1+2i & 0\end{array} \right] \; + \; \dfrac{1}{2} \left[ \begin{array}{ccc} -3i & 1-i & -3+2i \\ -1-i & i & -1-2i \\ 3+2i & 1-2i & 0 \end{array} \right] \\ \; \\ \; \\ = \dfrac{1}{2} \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \ \ \ which \; is \; a \; zero \; matrix. \\ \; \\ \; \\ \; \\ Now, \; Q\;=\; \dfrac{1}{2}(A-A^{\theta}) \;= \; \dfrac{1}{2} \left[ \begin{array}{ccc}3i & -1+i & 3-2i\\1+i & -i & 1+2i \\-3-2i & -1+2i & 0\end{array} \right] \; \; - \; \; \dfrac{1}{2} \left[ \begin{array}{ccc} -3i & 1-i & -3+2i \\ -1-i & i & -1-2i \\ 3+2i & 1-2i & 0 \end{array} \right] \\ \; \\ \; \\ = \left[ \begin{array}{ccc}3i & -1+i & 3-2i\\1+i & -i & 1+2i \\-3-2i & -1+2i & 0\end{array} \right] \; which \; is \; a \; skew \; Hermitian \; matrix $