Prove $T$ is diagonalizable and find a basis $B$ of $V$ such that the matrix associated to $T$ be diagonal. Let $V=\mathbb{R}^3$ and $T:V\rightarrow V$ the linear operator defined by $T(x,y,z)=(3x-z,2x+4y+2z,-x+3z)$
Prove  $T$ is diagonalizable and find a basis $B$ of $V$ such that the matrix associated to $T$ be diagonal.
Let $B=\{e_1,e_2,e_3\}$ the canonical basis of $V$
this implies $T_{BB}=\begin{bmatrix}
3 &2  &-1 \\ 
 0&4  &0 \\ 
 1&2  &3 
\end{bmatrix}$ then by definition
$P_T(\lambda)=det(xI-T)=det(xI-T_{BB})=det\begin{bmatrix}
-3+x &-2  &1 \\ 
 0&-4+x  &0 \\ 
 -1&-2  &-3+x 
\end{bmatrix}=-40+34x-10x^2+x^3=(x-4)(x^2-6x+10)$
in consequence, the characteristic polynomial of $T$ is $(x-4)(x^2-6x+10)$

Then, $x_1=4$ is an eigenvalue. My question, exists other eigenvalue?

If not, then $T$ is diagonalizable, because we have an unique eigenvalue an that eigenvalue go to produce an unique eigenvector and that vector is linearly independent.

Find the basis $B$ can be a little difficult to me, can someone help me?

 A: if a n * n matrix has n different  eigenvalues then it is diagonalizable.
here you have to solve $$det(xI−T)=0$$ for x .
so you have $$ (-4+x)[(-3+x)(-3+x)-1]=0$$ 
$$ (-4+x)(X^2+9-6x-1)=0$$
$$(-4+x)(x^2-6x+8)=0$$
$$x=4,-2,4$$
now you have to find eigenspace, you have to show that T-4I span 2 dimension subspaces.
A: $4$ is the only real eigenvalue. The remaining two are $3\pm i$.
Regarding the basis for which $T$ is diagonal, think of the set of eigenvctors associated with these eigenvalues.
Hint: After change to basis $V$, $T$ becomes $V^{-1}TV$.
A: This is just to elaborate on shere's answer:
Clearly your matrix is wrong. The transformation T is represented as: 
$$T_{BB}=\begin{bmatrix}
 3 & 2 & -1 \\ 
 0 & 4 & 0  \\ 
-1 & 2 & 3 
\end{bmatrix}$$
Now, this means that the characteristic equation is got by:
$$
det \begin{bmatrix}
 x-3 & -2  & 1  \\ 
 0   & x-4 & 0  \\ 
 1   & -2  & x-3 
\end{bmatrix} = 0
$$
This gives us the equation:
$$x^3-10x^2+32x-32=0$$
$$(x-4)(x^2-6x+8)=0$$
$$(x-4)(x-2)(x-4)=0$$
Therefore, the eigenvalues of $T$ are $4$ and $2$. (shere got it wrong here)
This gives us three linearly independent eigenvectors for $T$:
$$[1\quad 0\quad -1],\; [0\quad 1\quad 0],\; and\; [1\quad -2\quad 1]$$
which together form a basis and thus, $T$ is diagonalizable.
