so I have to calculate (straight, showing all the steps):
$$\mathrm{div} [r \times a]$$ where $a$ is a conts and $r$ is a motion vector, so first:
$$[r \times a] = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ x & y & z \\ a_x & a_y & a_z \end{vmatrix} = \vec{i}ya_z +\vec{k}xa_y+\vec{j}za_x-\vec{k}ya_x -\vec{i}za_y-\vec{j}zx_a$$
known that:
$$\mathrm{div} = \frac{\partial}{\partial{x}} + \frac{\partial}{\partial{y}} + \frac{\partial}{\partial{z}}$$
what should I substitute to div defenition above from the determinant?
I'm afraid the divergence operator isn't quite what you said! The divergence operator takes as input a vector field, and returns a scalar field. By definition, $$ {\rm div}(F_1 \vec i + F_2 \vec j + F_3 \vec k) = \frac{\partial F_1}{\partial x} +\frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}. $$
In your example, $$ F_1 = ya_z - za_y, \ \ \ \ F_2 = za_x - xa_z, \ \ \ \ F_3 = xa_y - ya_x,$$ so you just need to substitute this into my formula.