With the use of this formula
$$d\omega(X_1, \dots, X_{r+1}) = \sum_{i=1}^{r}(-1)^{i+1}X_i\omega(x_1,\dots,\hat{X}_i,\dots,X_{r+1})+\sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],X_1,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{r+1}) $$ where the term under $\hat{}$ is removed, I want to get:
$$3d\sigma(X,Y,Z)=-\sigma([X,Y],Z).$$
For the specific case when $X,Y\in V_K(M)$ and $Z\in V(M)$, with $M$ a smooth manifold and $K$ is the characteristic distribution of $\sigma$. And $V_K(M)$ denote all vectors fields that are tangent to $K$.
With all the information of the problem, here is what I have done: the only Lie bracket that should remain is $[X,Y]$ as (for some reason) the Lie bracket of a vector field that is tangent to the other field is zero. And the same is true(?) for the two-forms.
This yields
$$3d\sigma(X,Y,Z)=X_3\sigma(X_1,X_2)-\sigma([X_1,X_2],X_3).$$
Obviously I have misunderstood something fundamental with both differential forms and Lie bracket. From this I can at least conclude that I lack the knowledge why a Lie bracket of a vector field tangent to the other field is zero. And why the same two vector fields in a two-form should be zero.
I would be grateful if someone could answer these questions of mine and perhaps guide me towards the correct reasoning for this problem.
If $X \in K_M$, then $\iota_X\sigma = 0$, so for any $Z \in V(M)$, $0 = (\iota_X\sigma)(Z) = \sigma(X, Z)$. Also note that $\sigma(Z, X) = -\sigma(X, Z) = 0$. So if $X \in K_M$ appears as either argument of $\sigma$, then that term is zero.
Note that
$$d\sigma(X, Y, Z) = X\sigma(Y, Z) - Y\sigma(X, Z) + Z\sigma(X, Y) - \sigma([X, Y], Z) + \sigma([X, Z], Y) - \sigma([Y, Z], X).$$
As $X, Y \in K_M$, any term in which $X$ or $Y$ appears as an argument is zero. The only term which doesn't is $\sigma([X, Y], Z)$. Therefore, $d\sigma(X, Y, Z) = -\sigma([X, Y], Z)$.
I believe the missing $3$ is due to a different convention for the exterior derivative.
Note that the distribution $K$ is involutive (closed under Lie bracket), and hence integrable, if and only if $d\sigma = 0$.