Let $R$ be a $3 \times 3$ proper orthogonal and normal matrix such that $RR^T = I$ and $det(R) = 1$, so basically a rotation matrix.
$R = \begin{bmatrix} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \\ \end{bmatrix}$
Is the following relation true?
$r_{13} = r_{32}r_{21} - r_{22}r_{31}$
On
Let $\{e_1, e_2, e_3\}$ be the canonical basis and $u$ the rotation whose matrix is $R$. The given relation is equivalent to
$$\langle u(e_3), e_1 \rangle = \langle u(e_1) \times u(e_2), e_1 \rangle$$
$u(e_1), u(e_2)$ are unit orthogonal vectors. $u$ being a rotation, $u(e_1) \times u(e_2)$ is a unit vector equal to $u(e_3)$. Therefore the given relation is true. The very true relation is
$$u(e_3)= u(e_1) \times u(e_2).$$
We have:
$$\mathbf{RR^T}=\mathbf{I} \ \iff \ \mathbf{R^{-1}}=\mathbf{R^T}\tag{1}$$
Remember now the classical formula
$$\mathbf {A} ^{-1}=\frac{1}{\det(\mathbf {A})}\operatorname {adj} (\mathbf {A} )\tag{2}$$
where $\operatorname {adj} (\mathbf {A})$ is the adjugate matrix of square matrix $\mathbf {A}$, defined as the transpose of its matrix of cofactors.
Taking $\mathbf {A}=\mathbf {R}$ in (2), we obtain, using (1), the identity:
$$\mathbf {R^{-1}}= \operatorname {adj} (\mathbf {R} )=\mathbf {R^T} \implies \mathbf {R}= (\operatorname {adj} \mathbf {R)^T} $$
Otherwise said, $\mathbf {R} $ is equal to the matrix of its cofactors (without need to transpose it), explaining in particular your identity.