The most generic answer: any time that we can reduce a problem over an incredibly general object (say, a matrix) to a problem in which we have more information at our fingertips (say, the same problem but over matrices that are in JCF), we make life easier - both in terms of proving theory and in terms of practical computations.
To be more specific to the situation at hand: the Jordan canonical form is sort of the next-best-thing to diagonalization. If the matrix is diagonalizable, then its JCF is diagonal; if it isn't, then what you get is at least block diagonal, and the blocks come in a predictable form.
We have:
$$A = \begin{bmatrix}1&1&1\\0&2&0 \\ 0&0&2\end{bmatrix}$$
We find the eigenvalues of $|A - \lambda I| = 0$, hence:
$$\lambda_1 = 1, \lambda_{2,3} = 2$$
That is, we have a single root and a double root eigenvalue, algebraic multiplicity.
To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$.
For $\lambda_1 = 1$, we get the eigenvector:
$$v_1 = (1,0,0)$$
For $\lambda_{2,3} = 2$, we get the eigenvectors (normally, we do not get two linearly independent eigenvectors):
$$v_2 = (1,0,1), v_3 = (1,1,0)$$
We now can write $P$ using the eigenvectors as columns. We have,
$$P = [v_1 | v_2 | v_3 ] = \begin{bmatrix}1&1&1\\0&0&1 \\ 0&1&0\end{bmatrix}$$
We can write the Jordan Normal Form (notice that we do not have any Jordan blocks), $J$, using the corresponding eigenvalues:
$$J = P^{-1} A P$$
However, we can also write this straight off from the eigenvalues and knowing we do not need any Jordan blocks.
$$J = \begin{bmatrix}1&0&0\\0&2&0 \\ 0&0&2\end{bmatrix}$$
Lastly, we should verify:
$$A = P J P^{-1}$$
I purposely left things so you can fill in the details of the calculations.
Best Answer
A matrix admits a Jordan canonical form just when all its eigenvalues lie in the base field. (See https://en.wikipedia.org/wiki/Jordan_normal_form).
So all you need is a matrix with at least one nonreal eigenvalue. Write down the matrix for a quarter circle rotation in the plane.