= Solution
Compute the <characteristic polynomial> using $\det(\lambda I-M)$:
$$
\det(\lambda I-M)=\lambda^3-7\lambda^2+6
=(\lambda-1)(\lambda^2-6\lambda-6).
$$
Its three distinct real roots are $1$ and $3\pm\sqrt{15}$. For $\lambda=1$, the first row of $(M-I)v=0$ gives $z=3y$ and the second then gives $x=y$, so take $v_1=(1,1,3)^T$.
For either root of $\lambda^2-6\lambda-6=0$, take $z=1$ and solve the row equations to get $v=(\lambda-8,5-\lambda,1)^T$. Substitution makes the first two row residuals opposite multiples of the quadratic equation and the third zero. Thus a complete choice is
$$
\boxed{\begin{aligned}
\lambda_1&=1,&v_1&=(1,1,3)^T,\\
\lambda_-&=3-\sqrt{15},&v_-&=(-5-\sqrt{15},\ 2+\sqrt{15},\ 1)^T,\\
\lambda_+&=3+\sqrt{15},&v_+&=(-5+\sqrt{15},\ 2-\sqrt{15},\ 1)^T.
\end{aligned}}
$$
Every nonzero <scalar> multiple is another valid <eigenvector> in the corresponding one-dimensional <eigenspace>.
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