An eigenvector $\mathbf{x}=\left[\begin{array}{l}x \\ y\end{array}\right]$ is one that $A$ maps to a multiple of itself, that is, there's some number $\lambda$ so that $A \mathbf{x}=\lambda \mathbf{x}$. Let our matrix be $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$. Then the matrix equation $A \mathbf{x}=\lambda \mathbf{x}$ says
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
\lambda x \\
\lambda y
\end{array}\right]
$$
One reason that eigenvectors are important is that you can use them to understand what the transformation is doing geometrically.
If the rows and columns in $A$ each add to the same number $k$, then $a+b=a+c=b+d$, so $b=c$, and $a=d$. So our matrix looks like $A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]$, and then our equation looks like
$$
\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
\lambda x \\
\lambda y
\end{array}\right]
$$
which we can write as two equations
$$
\begin{aligned}
& a x+b y=\lambda x \\
& b x+a y=\lambda y
\end{aligned}
$$
Now, are any of the three named vectors eigenvectors? Just the vector III. That works because the two equations each say $a+b=\lambda$ which tells you what the constant $\lambda$ is. So the answer is (C).
