Given, $A=\left[\begin{array}{lll}2 & 1 & 1 \\ 2 & 3 & 2 \\ 3 & 3 & 4\end{array}\right]$

So, $|A-\lambda I|=0$ ,

which is the Characteristic Equation, and I is the unit matrix of order 3.

$\Rightarrow\left|\begin{array}{ccc}2-\lambda & 1 & 1 \\ 2 & 3-\lambda & 2 \\ 3 & 3 & 4-\lambda\end{array}\right|=0$

Solving via First row :

$\Rightarrow(2-\lambda)\{(3-\lambda)(4-\lambda)-6\}-\{8-2 \lambda-6)+\{6-9+3 \lambda\}$

$\Rightarrow \quad(2-\lambda)(\lambda-6)(\lambda-1)+2(\lambda-1)+3(\lambda-1)=0$

$\Rightarrow \quad(\lambda-1)\{(2-\lambda)(\lambda-6)+5\}=0$

$\Rightarrow \quad-(\lambda-1)\left(\lambda^{2}-8 \lambda+7\right)=0$

$\Rightarrow \quad(\lambda-1)(\lambda-7)(\lambda-1)=0$

Hence, Eigen values are $\rightarrow 1,7$ and 1 .

Now, for Eigen vectors :

$$ \left[\begin{array}{ccc} 2-\lambda & 1 & 1 \\ 2 & 3-\lambda & 2 \\ 3 & 3 & 4-\lambda \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=0 $$ At, $\lambda=7$ $$ \left[\begin{array}{ccc} -5 & 1 & 1 \\ 2 & -4 & 2 \\ 3 & 3 & -3 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=0 $$

$$ \begin{array}{r} \Rightarrow-5 x+y+z=0 \\ 2 x-4 y+2 z=0 \\ 3 x+3 y-3 z=0 \end{array} $$ Solving first two equations;

\begin{array}{l} \Rightarrow \quad \frac{x}{2-(-4)}=\frac{y}{2-(-10)}=\frac{z}{20-2} \ \Rightarrow \quad \frac{x}{6}=\frac{y}{12}=\frac{z}{18} \end{array} $$ $$ $$\Rightarrow \frac{x}{1}=\frac{y}{2}=\frac{z}{3} $$

Thus, one eigen vector is [1, 2, 3].

At, $\lambda=1$ $$ \left[\begin{array}{lll} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=0 $$ $$ \begin{array}{l} \Rightarrow x+y+z=0 \\ 2 x+2 y+2 z=0 \\ 3 x+3 y+3 z=0 \end{array} $$

Since, above all three equations are same,

Let , y = m ; z = n ;

then, x = - (m + n). (since, x+y+z = 0)

At, m = 0 , n = 1 , x = -(0 + 1) = -1

At, m = 1 , n = 0 , x = -(1 + 0) = -1

Thus, other two eigen vectors are [-1, 0, 1] and [-1, 1, 0] respectively.