Answer
See below
Work Step by Step
Given
$\begin{bmatrix}
2-\lambda & -2 & 3\\
1 & -1-\lambda & 3 \\
1 & -2 & 4-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\lambda_1=1,\lambda_2=3$
a) For $\lambda=1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1 & -2 & 3\\
1 & -2 & 3 \\
1 & -2 & 3
\end{bmatrix}$
Let $r,s$ be free variables.
$\vec{V}=r(2,1,0)+s(-3,0,1) \\
E_1=\{(2,1,0);(-3,0,1)\}
\rightarrow dim(E_1)=2$
Let $v_1=u_1=(-3,1,0)\\
v_2=u_2=(2,1,0)$
Use the Gram-Schmidt procedure:
$u_2=v_2-\frac{}{||u_1||^2} u_1=(2,1,0)+\frac{6}{10}(-3,0,1)=\frac{1}{5}(1,5,3)$
An orthogonal basis for $E_1=\{(-3,0,1);(1,5,3)\}$
b) $\lambda=3$
$B=\begin{bmatrix}
-1 & -2 & 3\\
1 & -4 & 3 \\
1 & -2 & 1
\end{bmatrix}$
Let $t$ be free variables.
$\vec{V}=t(1,1,1) \\
E_2=\{(1,11)\}
\rightarrow dim(E_1)=1$
Let $v=(1,1,1)$
Obtain $=-2 \ne 0\\
=\frac{9}{5} \ne 0$
The vectors in $E_1$ are not orthogonal to the vectors in
$E_2$
