How to evaluate the value of $\begin{bmatrix}\vec{l},\vec{m},\,\vec{n}\end{bmatrix} \begin{bmatrix}\vec{a},\vec{b},\,\vec{c}\end{bmatrix}$ 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. 
 A: 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.
