can two eigenvectors be same for a triangular matrix whose answer is have 3 eigenvalues which are same so the matrix is 1  4  2
                0  1  6
                0  0  1
the eigenvalues are 1,1,1
and i am getting eigenvectors :- [0,-1,2], and [1,0,0], [1,0,0]. 
can eigenvectors be same ?  if not can you please tell me where am i wrong ?

How the vector [0,-1,2] was produced:
we get 2 equations , 
4x2 + 2x3 =0 ;
x3= 0 ;
4x2 = -2x3 ==> 2x2 = -x3
 now we cross multiply -1 and 2 
we get
 x2/2 = x3/-1 
then we were told  that x2 will be 2 and x3 will be -1
this is how i got the 2nd eigenvector
 A: Let your matrix be $A$; then it's clear that the rank of $A-I$ is $2$, hence the nullity is $1$.
The eigenvector should be $(1,0,0)^T$.
Check that $(0,-1,2)^T$ is not an eigenvector.
Remark about your working to obtain $(0,1,-2)$ as an eigenvector.
From, $4x_2+2x_3=0$ and $x_3=0$, that would force $x_2=0$ as well. Notice that $(x_2,x_3)=(0,0)$ satisfies $2x_2=-x_3$ as well but $(0,-1,2)$ doesn't satisfy the equations.
A: You should not have a case where eigenvectors are repeated and besides, for your matrix, you would only get the eigenvector $(1,0,0)^T$. I am not sure how you got $(0,-1,2)^T$. This also implies that your matrix is not diagonalizable.
A: An eigenvector algorithm typically "tries" to produce a complete and linearly independent set of eigenvectors.  However, you are getting repeating eigenvectors because it is impossible to do this in your case since your matrix is not diagonalizable.
The vector $(1,0,0)$ is indeed an eigenvector of your matrix.  It is not clear to me why the vector $(0,-1,2)$ was produced.

Regarding your latest edit:
Your matrix was 
$$
A = \pmatrix{1 & 4 & 2\\
0 & 1 & 6\\
0 & 0 & 1}.
$$
As you have correctly indicated, the way we find the eigenvectors for $\lambda = 1$ (the only eigenvector in this case) is by solving the system given by $(A - I)x = 0$.  This produces the system of equations
$$
4x_2 + 2x_3 = 0\\
6x_3 = 0.
$$
You then correctly found the eigenvector $(1,0,0)$ by noting that the only solutions to this system are given by $x_2 = x_3 = 0$.  This presented a problem, however, since you expected 3 eigenvectors.
The correct solution at this point is to simply say that we can only produce 1 eigenvector, so the matrix is not diagonalizable.
The incorrect approach you attempted was to look for another solution to the system, with the thought that if $x$ solves only the first line then maybe that will be "good enough".  This is not the case: in order for $x = (x_1,x_2,x_3)$ to be an eigenvector, it needs to solve the full system of equations.
