Characteristic equation is

$|A - \lambda I| = 0 \\ \begin{vmatrix} 4 - \lambda & 6 & 6 \\ 1 & 3 - \lambda & 2 \\ -1 & -5 & -2 - \lambda \end{vmatrix} = 0 \\ (4 - \lambda)[(3 - \lambda) (-2-\lambda) - 2(-5)] - 6(-2 - \lambda - 2(-1)) + 6(-5 -(-1)(3-\lambda)) = 0 \\ (4 - \lambda) [-6 - 3\lambda + 2\lambda + \lambda^2 + 10] - 6(-2 - \lambda + 2) + 6(-5 - (-3 + \lambda)) = 0 \\ (4 - \lambda) (\lambda^2 - \lambda + 4) + 6\lambda + 6(-5 + 3 - \lambda) = 0 \\ (4 - \lambda) (\lambda^2 - \lambda + 4) + 6\lambda + 6(-2 - \lambda) = 0 \\ 4\lambda^2 - 4\lambda + 16 - \lambda^3 + \lambda^2 + 6\lambda - 12 - 6\lambda = 0 \\ -\lambda^3 + 5\lambda^2 - 8\lambda + 4 = 0 \\ \lambda^3 + 5\lambda^2 + 8\lambda - 4 = 0 \\ \lambda = 1,2,2$

Eigen values are 1, 2, 2

Let X be eigen vector for $\lambda$ = 1

$(A - 1I)X = 0 \\ \begin{bmatrix} 3 & 6 & 6 \\ 1 & 2 & 2 \\ -1 ^ -5 & -3 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} + R_{2} \begin{bmatrix} 3 & 6 & 6 \\ 1 & 2 & 2 \\ 0 & -3 & -1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} \leftrightarrow R_{2} \\ \frac{R_{1}}{2} \begin{bmatrix} 3 & 6 & 6 \\ 1 & 2 & 2 \\ 0 & -3 & -1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $

Rank = 2

No. of unknown = 3

$\therefore$ 3 - 2 = 1 linear independent solution

$-3x_{2} - x_{3} = 0$

Let $x_{2} = 1 \\ -3 - x_{3} = 0 \\ x_{3} = -3 \\ x_{1} + 2x_{2} + 2x_{3} = 0 \\ x_{1} + 2x_{2} - 6 = 0 \\ x_{1} + 2 - 6 = 0\\ x_{1} = 0$

$X = \begin{bmatrix} 4 \\ 1 \\ -3 \end{bmatrix} \rightarrow$ Eigen vector for $\lambda$ = 1

Lwt Y be eigen vector for $\lambda$ = 2

$(A - 2I)Y = 0 \\ \begin{bmatrix} 2 & 6 & 6 \\ 1 & 1 & 2 \\ -1 & -5 & -4 \end{bmatrix} \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \frac{R_{1}}{2} \\ R_{3} + R_{2} \\ R_{2} = R_{1} \begin{bmatrix} 1 & 3 & 3 \\ 0 & -2 & -1 \\ 0 & -4 & -2 \end{bmatrix} \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ R_{3} - 2R_{2} \begin{bmatrix} 1 & 3 & 3 \\ 0 & -2 & -1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} y_{1} \\ y_{2} \\ y_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Rank = 2

No. of unknown = 3

$\therefore$ 3 - 2 = 1 linear independent solution

$-2y_{2} - y_{3} = 0$

Let $y_{2} = 1 \\ -2 - y_{3} = 0 \\ y_{3} = -2 \\ y_{1} + 3y_{2} + 3y_{3} = 0 \\ y_{1} + 3 - 6 = 0 \\ y_{1} - 3 = 0 \\ y_{1} = 3$

$Y = \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix} \rightarrow$ Eigen vector for $\lambda$ = 2