The eigenvalues are $1, a, -a$ so if $a^2 \neq 1,$ then $A$ has three different eigenvalues so $A$ is diagonalizable.
Case $a^2 = 1$ :
the rank of $$A - I = \pmatrix{0&0&0\\1&-1&1\\1&1&-1}$$ is $2$ because the first and second columns are linearly independent, so the dimension of the null space of $A-I$ is $1$. Therefore when $a^2=1, A$ is not diagonalizable.
added after the user1551's comment.
Case $a = 0$:
the rank of $A$ is $2$ therefore the null space of $A$ has dimension $1$ so again $A$ is not diagonalizable.
First note that if $k\neq 2,3$, then all eigenvalues are different, and hence the matrix is diagonalizable.
In case when $k = 2$ or $k=3$, i.e. you have an eigenvalue with multiplicity more than $1$, you need to calculate dimension of the eigenspace. If all the dimensions of eigenspaces are equal to multiplicities of corresponding eigenvalues, then the matrix is diagonalizable, otherwise not.
Can you continue from here?
Edit: Let $\lambda$ be an eigenvalue of $A$ and $E_\lambda$ its corresponding eigenspace. Let $p_A$ be the characteristic polynomial of $A$.
Since $\lambda$ is an eigenvalue, $p_A(\lambda) = 0$ and thus, there exists positive integer $n$ such that $p_A(x) = (x-\lambda)^nq(x)$ and $q(\lambda)\neq 0$. Define $a_\lambda$ as this $n$ in the corresponding factorization of $p_A$. We call this algebraic multiplicity of eigenvalue $\lambda$.
On the other hand, we are also interested in number $g_\lambda = \dim E_\lambda = \dim\ker(A-\lambda I)$ which is called geometric multiplicity of eigenvalue $\lambda$.
In general, $1\leq g_\lambda \leq a_\lambda$ and $A$ is diagonalizable if and only if $a_\lambda = g_\lambda$ for all eigenvalues $\lambda$ of $A$.
In your case, you have characteristic polynomial $p_A(x) = (x-2)(x-3)(x-k)$. If $k\neq 2,3$, then $a_2 = a_3 = a_k = 1$ and since $1\leq g_\lambda \leq a_\lambda$, we have $g_2=g_3 = g_k = 1$, as well. That is, all geometric and algebraic multiplicities are equal and the matrix is diagonalizable.
In the case $k = 3$, we have $a_2 = 1$, $a_3 = 2$, so again, $a_2 = g_2 = 1$, but we need to check whether $g_3 = 1$ or $g_3 = 2$. To find $g_3$ solve the homogeneous linear system $(A-3I)v = 0$. You will find out that solution space is of dimension $2$ (its obvious as soon as you write it), and hence $a_3 = g_3 = 2$, so $A$ is diagonalizable.
In the case $k = 2$, similar to the previous case, we have $a_3 = g_3 = 1$, but $a_2 = 2$ so we need to solve homogeneous linear system $(A-2I)v = 0$, but in this case you will find out that $g_2 = 1 \neq a_2$, and thus $A$ is not diagonalizable.
Best Answer
Right, we can read off the characteristic polynomial from the diagonal: $$p(\lambda) = (\lambda-1)(\lambda-2)^2.$$Call the matrix $A$. If the matrix is to be diagonalizable, we must obtain a basis of eigenvectors. We need three vectors. In other words, since $$\dim(\ker(A-{\rm Id})) \geq 1, \quad \dim(\ker(A-2\,{\rm Id})) \geq 1,$$ and $1$ is a simple root of $p(\lambda)$, we have $\dim(\ker(A-{\rm Id}))=1$. So $\dim(\ker(A-2\,{\rm Id}))$ had better be $2$. $$A-2\,{\rm Id } = \begin{bmatrix} -1 & 0 & 1 \\ 0 & 0 & k \\ 0 & 0 & 0 \\ \end{bmatrix}$$ From this it is clear that we get what we want for $k=0$.