Why did you change matrices? The matrix in the attachment is the same one you mentioned first:
\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]
but you asked how to find [1 0 0] as an eigenvalue of
\left[\begin{array}{ccc}2 & 2 & 3 \\0 & 6 & 6 \\ 0 & 6 & 6\end{array}\right]
and it isn't! Did you miscopy?
I think people may be trying to give you too "sophisticated" answers when your questions are really very simple. I must say I found your original question very peculiar! you had done the hard part, found that the eigenvalue equation was (\lambda- 1)((\lambda- 5)^2- 36)= 0 and apparently were asking how to solve that equation! If you are doing linear algebra you should have learned how to solve polynomial equations, especially those that are already factored, long ago! In order to have that product, (\lambda- 1)((\lambda- 5)^2- 36) equal to 0, you must have either \lambda- 1= 0 or (\lambda-5)^2- 36 equal to 0. From \lambda- 1= 0 we get, of course, \lamba= 1. From (\lambda- 5)^2- 36= 0 we get (\lambda- 5)^2= 36 so \lambda- 5= \pm 6. \lambda= 5+ 6= 11 or \lambda= 5- 6= -1.
Now to find the corresponding eigenvectors, just do the computation:
For \lambda= 1, for example,
\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]\left[\begin{array}{c}a \\ b \\ c\end{array}\right]= \left[\begin{array}{c} 1*a \\ 1*b \\ 1*c\end{array}\right]
Multiply that out and you get 3 equations for a, b, c:
a+ 2b+ 3c= a, 5b+ 6c= b, and 6b+ 5c= c, or 2b+ 3c= 0, 4b+ 6c= 0, 6b+ 4c= 0. Three independent equations in 3 variables will have ONE solution: obviously a= b= c= 0 is a solution. Since 1 is an eigenvalue, this must have non-trivial solutions and so an infinite number of solutions: the equations must not be independent. In fact, a has disappeared from the equations so a can be anything. Notice that 4b+ 6c= 2(2b+ 3c) so 2b+ 3c= 0 and 4b+ 6c= 0 are really the same equation. But we can still solve 2b+ 3c= 0 and 6b+ 4c= 0, to get b= c= 0. Any eigenvector is of the form [a 0 0]= a[1 0 0].
Similarly for the other eigenvalues. If \lamba= -1 then the matrix equation becomes
\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]\left[\begin{array}{c}a \\ b \\ c\end{array}\right]= \left[\begin{array}{c} -1*a \\ -1*b \\ -1*c\end{array}\right]
Multiplying that out gives a+ 2b+ 3c= -a, 5b+ 6c= -b, and 6b+ 5c= -c, or 2a+ 2b+ 3c= 0, 6b+ 6c= 0, 6b+ 6c= 0. (Was that where you got that matrix before?) Obviously the last two equations are the same: b= -c. Putting that into the first equation, 2a+ 2b- 3b= 0 gives b= 2a. Any eigenvector, corresponding to eigenvalue -1 is of the form [a, 2a, -2a]=
a[1, 2, -2].
For \lambda= 11 it is
\left[\begin{array}{ccc}1 & 2 & 3 \\0 & 5 & 6 \\ 0 & 6 & 5\end{array}\right]\left[\begin{array}{c}a \\ b \\ c\end{array}\right]= \left[\begin{array}{c} 11*a \\ 11*b \\ 11*c\end{array}\right]
Multiplying, a+ 2b+ 3c= 11a, 5b+ 6c= 11b, and 6b+ 5c= 11c, or -10a+ 2b+ 3c= 0, -6b+ 6c= 0, 6b- 6c= 0. Again, the last two equations are equivalent and give b= c. Putting that into the first equation, -10a+ 2b+ 3b= 0 gives -10a+ 5b= 0 or b= -2a. Any eigenvector corresponding to \lambda= 11 must be of the form [a, -2a, -2a]= a[1, -2, -2].
To answer your last question, "normalizing" a vector means reducing it to unit length: find the length of the vector and divide each component by that. [1 0 0] already has length 1 and is already normalized. [1 2 -2] has length \sqrt{1^2+ 2^2+(-2)^2}= \sqrt{9}= 3. The unit vector in the direction of [1 2 -2] is \left[\begin{array}{ccc}\frac{1}{3} & \frac{2}{3} & -\frac{2}{3}\end{array}\right]. [1 -2 -2] also has length 3 and a unit vector in its direction is \left[\begin{array}{ccc}\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3}\end{array}\right].