Find the Eigenvector of a matrix Find the eigenvectors of the matrix

$$\displaystyle\begin{bmatrix}
0 &amp;2  &amp;3 \\ 
-2 &amp;0  &amp;5 \\ 
-3 &amp;-5  &amp;0 
\end{bmatrix}.$$ 

So I start with $|A-\lambda I|=0$

$$\begin{bmatrix}
-\lambda &amp;2  &amp;3 \\ 
-2 &amp;-\lambda  &amp;5 \\ 
-3 &amp;-5  &amp;-\lambda 
\end{bmatrix}=0$$

$$-\lambda(\lambda^2+25)-2(2\lambda+15)+3(10-3\lambda)=0 $$

$$\lambda^3+38\lambda=0$$

The given eigenvalue is -2, which is later used for finding the eigenvector. My equation won't give -2. This is also a part of a multiquestion series, so I cannot think that the question has something wrong in it.

What did I do wrong here ?
Your characteristic polynomial is right, it is $\lambda^3 + 38 \lambda = 0$. But $-2$ is not a root of it, since $(-2)^3 + 38 \cdot (-2) = -84$. If you transform the equation like this $\lambda^3 + 38 \lambda = \lambda ( \lambda^2+38)=0$, then you find that $\lambda_1 = 0$ is a real eigenvalue, and there are two complex $\lambda_{2,3} = \pm i \sqrt{38}$. I let you the rest to solve.
You have not done anything wrong. I have checked your
answer with gp (available from http://pari.math.u-bordeaux.fr/).
I am awful in numerical calculations!
Check whether 
you have copied the matrix correctly.