 Solution

Let $A_1,A_2,A_3$ denote the three matrices in their printed order. A Hermitian matrix satisfies $A^{†}=A$, a unitary matrix satisfies $A^{†}A=I$, and an orthogonal matrix is real with $A^{T}A=I$.

For the first matrix, $A_1^{†}=A_1$ and $A_1^2=I$. It is therefore Hermitian and unitary. It is not real, and even allowing complex entries in the transpose test gives $A_1^T{A}_1=-I$, so it is not orthogonal.

The second matrix is $A_2=iI$. Thus $A_2^{†}{A}_2=(-iI)(iI)=I$, whereas $A_2^{†}=-A_2$ and $A_2^T{A}_2=-I$. It is unitary only among the listed classes.

The third matrix is real, but direct multiplication givesIt is neither orthogonal nor unitary. Its unequal off-diagonal transpose entries also show $A_3^T\ne A_3$, so it is none of the three. The common prefactor $1/2$ applies to the central diagonal entry as well; replacing that entry by one would describe a different matrix.