Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be three mutually orthogonal unit vectors, such that
\[\mathbf{a} = p (\mathbf{a} \times \mathbf{b}) + q (\mathbf{b} \times \mathbf{c}) + r (\mathbf{c} \times \mathbf{a})\]for some scalars $p,$ $q,$ and $r,$ and $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1.$  Find $p + q + r.$
Taking the dot product of the given equation with $\mathbf{a},$ we get
\[\mathbf{a} \cdot \mathbf{a} = p (\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})) + q (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})) + r (\mathbf{a} \cdot (\mathbf{c} \times \mathbf{a})).\]Since $\mathbf{a}$ is orthogonal to both $\mathbf{a} \times \mathbf{c}$ and $\mathbf{c} \times \mathbf{a},$ we are left with
\[\mathbf{a} \cdot \mathbf{a} = q (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})) = q.\]Then $q = \mathbf{a} \cdot \mathbf{a} = 1.$

Similarly, if we take the dot product of the given equation with $\mathbf{b},$ we get
\[\mathbf{b} \cdot \mathbf{a} = p (\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b})) + q (\mathbf{b} \cdot (\mathbf{b} \times \mathbf{c})) + r (\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})).\]Since $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, we are left with
\[0 = r (\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})).\]By the scalar triple product, $\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1,$ so $r = 0.$  Similarly, by taking the dot product of both sides with $\mathbf{c},$ we are left with $p = 0.$

Therefore, $p + q + r = \boxed{1}.$
Taking the dot product of the given equation with $\mathbf{a},$ we get
\[\mathbf{a} \cdot \mathbf{a} = p (\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})) + q (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})) + r (\mathbf{a} \cdot (\mathbf{c} \times \mathbf{a})).\]Since $\mathbf{a}$ is orthogonal to both $\mathbf{a} \times \mathbf{c}$ and $\mathbf{c} \times \mathbf{a},$ we are left with
\[\mathbf{a} \cdot \mathbf{a} = q (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})) = q.\]Then $q = \mathbf{a} \cdot \mathbf{a} = 1.$

Similarly, if we take the dot product of the given equation with $\mathbf{b},$ we get
\[\mathbf{b} \cdot \mathbf{a} = p (\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b})) + q (\mathbf{b} \cdot (\mathbf{b} \times \mathbf{c})) + r (\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})).\]Since $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, we are left with
\[0 = r (\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})).\]By the scalar triple product, $\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1,$ so $r = 0.$  Similarly, by taking the dot product of both sides with $\mathbf{c},$ we are left with $p = 0.$

Therefore, $p + q + r = \boxed{1}.$
The final answer is $The final answer is $\boxed{1}$.$. I hope it is correct.