= Solution
The second coordinate direction is already an <eigenvector> with <eigenvalue> three. The first and third coordinates form a symmetric block, whose sum and difference directions give
$$
\boxed{\begin{array}{c|c}
\text{eigenvalue}&\text{unit eigenvector}\\\hline
6&(1,0,1)^T/\sqrt2\\
3&(0,1,0)^T\\
2c-4&(1,0,-1)^T/\sqrt2
\end{array}.}
$$
For example, the sum of the block's diagonal and off-diagonal entries is $(1+c)+(5-c)=6$, while their difference is $(1+c)-(5-c)=2c-4$. Equivalently the characteristic polynomial is $(\lambda-6)(\lambda-3)(\lambda-2c+4)$.
The displayed basis works for every real $c$. At $c=7/2$, the <eigenvalue> three has <eigenspace> spanned by the second and third listed vectors; at $c=5$, the <eigenvalue> six has <eigenspace> spanned by the first and third. Every nonzero vector in a repeated <eigenspace> is an <eigenvector>, and any <orthonormal basis> of it may replace the displayed choices.
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