There exists a scalar $k$ such that for any vectors $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ such that $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0},$ the equation
\[k (\mathbf{b} \times \mathbf{a}) + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}\]holds.  Find $k.$
Since $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0},$ $\mathbf{c} = -\mathbf{a} - \mathbf{b}.$  Substituting, we get
\[k (\mathbf{b} \times \mathbf{a}) + \mathbf{b} \times (-\mathbf{a} - \mathbf{b}) + (-\mathbf{a} - \mathbf{b}) \times \mathbf{a} = \mathbf{0}.\]Expanding, we get
\[k (\mathbf{b} \times \mathbf{a}) - \mathbf{b} \times \mathbf{a} - \mathbf{b} \times \mathbf{b} - \mathbf{a} \times \mathbf{a} - \mathbf{b} \times \mathbf{a} = \mathbf{0}.\]Since $\mathbf{a} \times \mathbf{a} = \mathbf{b} \times \mathbf{b} = \mathbf{0},$ this reduces to
\[(k - 2) (\mathbf{b} \times \mathbf{a}) = \mathbf{0}.\]We must have $k = \boxed{2}.$