$\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot ( \vec{c} \times \vec{a})$?

I've seen my teacher of general physics write $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot ( \vec{c} \times \vec{a})$ but I've search for a proof in google and there is nothing. I also tried to prove it myself but I couldn't. I need help

One of the many ways to define the scalar triple product is: Let $\vec{a}=(a_1,a_2,a_3)$, $\vec{b}=(b_1,b_2,b_3)$ and $\vec{c}=(c_1,c_2,c_3)$ then we have $$ \vec{a}\cdot (\vec{b}\times\vec{c})=\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix} $$ So exchanging $f_1\leftrightarrow f_2$ and $f_2\leftrightarrow f_3$ we have this $$ \begin{array}{rcr} \vec{a}\cdot (\vec{b}\times\vec{c})&=&\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix}\\ &=&-\begin{vmatrix} b_1&b_2&b_3\\ a_1&a_2&a_3\\ c_1&c_2&c_3\\ \end{vmatrix}\\ &=&(-1)^2\begin{vmatrix} b_1&b_2&b_3\\ c_1&c_2&c_3\\ a_1&a_2&a_3\\ \end{vmatrix}&=\vec{b}\cdot (\vec{c}\times\vec{a}) \end{array} $$

The result follows from the identity $a\cdot (b\times c) = \det(a|b|c)$, where the determinant is of the matrix whose columns (or rows, if you prefer) are $a$, $b$, and $c$ respectively. Then, you need to know that swapping two columns of a matrix will multiply the determinant by $-1$. This yields:

$$\begin{split}a\cdot (b\times c) &= \det(a|b|c) \\ &= -\det(b|a|c) \\ &= \det(b|c|a)\\ &= b\cdot (c\times a)\end{split}.$$

As for why we have the relationship to determinants, this follows from expansion by minors for computing determinants, and the definition of cross product. In some sense, making this formula hold is why the cross product is defined like it is.