Lets the value of $\,\begin{bmatrix}\vec{l},\vec{m},\,\vec{n}\end{bmatrix}$ is $\,\vec{l}.\left(\vec{m}\times\vec{n}\right)$. We have to show that $$ \begin{bmatrix}\vec{l},\vec{m},\,\vec{n}\end{bmatrix} \begin{bmatrix}\vec{a},\vec{b},\,\vec{c}\end{bmatrix} = \begin{bmatrix} \vec{l}.\vec{a} & \vec{l}.\vec{b} & \vec{l}.\vec{c} \\ \vec{m}.\vec{a} & \vec{m}.\vec{b} & \vec{m}.\vec{c} \\ \vec{n}.\vec{a} & \vec{n}.\vec{b} & \vec{n}.\vec{c} \end{bmatrix}$$
How can I show this? Any advice is of great help.
Recall that the product of determinants of two $n\times n$ matrices is equal to the determinant of the product of these matrices: $$ \det\left(A\right)\det\left(B\right) = \det\left(AB\right), $$ and that the determinant of a matrix is equal to the determinant of its transpose:
\begin{align} \det\left(A\right) &= \det\left(A^T\right) &\implies&& \det\left(A\right)\det\left(B\right) &= \det\left(AB^T\right) \end{align}
Then we can write
\begin{align} \begin{bmatrix}\vec{l},\vec{m},\,\vec{n}\end{bmatrix} \begin{bmatrix}\vec{a},\vec{b},\,\vec{c}\end{bmatrix} & = \begin{vmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{vmatrix} \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{vmatrix} \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \\ &= \begin{vmatrix} \big\langle \vec{l}, \vec{a} \big\rangle & \big\langle \vec{l}, \vec{b} \big\rangle & \big\langle \vec{l}, \vec{c} \big\rangle \\ \big\langle \vec{m}, \vec{a} \big\rangle & \big\langle \vec{m}, \vec{b} \big\rangle & \big\langle \vec{m}, \vec{c} \big\rangle \\ \big\langle \vec{n}, \vec{a} \big\rangle & \big\langle \vec{n}, \vec{b} \big\rangle & \big\langle \vec{n}, \vec{c} \big\rangle \end{vmatrix} \end{align} Q.E.D.