<s>[INST] Determine the entries of the matrix given vectors (1,1,1), (1,0,-1), and (1,-1,0) are eigenvectors of the following matrix M =$\left[\begin{array}{ccc}1& 1 & 1 \\a & b & c \\d & e & f\end{array}\right]$. 
Work so far:
Let $\lambda$ denote the eigenvalues. By my calculations, the eigenvalues must $\lambda = 3,0,0$.
$M_1 =\left[\begin{array}{ccc}-2 & 1 & 1 \\a & b-3 & c \\d & e & f-3\end{array}\right]$.$M_1*x = (3,3,3)^T$.
$M_2 =\left[\begin{array}{ccc}1& 1 & 1 \\a & b & c \\d & e & f\end{array}\right]$. $M_2x = (0,0,0)^T$.
I'm not sure where to go from here. 
 [/INST] Note that since $\left [ \begin{array}{ccc}
1&1&1 \\
\end{array} \right ]^T$ is an eigenvector of the above matrix you must have that
$$
\left [ \begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
d & e & f \\
\end{array} \right ] \left[ \begin{array}{c}
1 \\
1 \\
1 \\
\end{array} \right ] \;\; =\;\; 
\left [ \begin{array}{c}
3 \\
a + b + c \\
d + e + f \\
\end{array} \right ] \;\; =\;\; \lambda\left [ \begin{array}{c}
1\\
1\\
1\\
\end{array} \right ]
$$
hence $\lambda = 3 = a+b+ c = d  +e  +f$ is one eigenvalue.  Applying the same idea to the second vector $\left [ \begin{array}{ccc}
1 & 0 & -1 \\
\end{array} \right ]^T$ we obtain
$$
\left [ \begin{array}{c}
0 \\
a - c \\
d - f \\
\end{array} \right ] \;\; =\;\; \lambda\left [ \begin{array}{c}
1 \\
0 \\
-1 \\
\end{array} \right ]
$$
so $\lambda = 0$ is another eigenvalue constraining $a = c$ and $d = f$.  Can you follow the pattern now?
</s>