If $\vec a,\vec b,\vec c$ are three non-zero non-null vectors and $\vec r$ is any vector in space, then $[\vec b\,\,\vec c\,\,\vec r]\vec a+[\vec c\,\,\vec a\,\,\vec r]\vec b+[\vec a\,\,\vec b\,\,\vec r]\vec c$ is equal to

$[\vec a\,\,\vec b\,\,\vec c]\vec r$

We have, $\vec r = x\vec a+y\vec b +z\vec c$   ...(i)

Taking product successively with $\vec b×\vec c, \vec c×\vec a$ and $\vec a ×\vec b$, we obtain

$x=\frac{[\vec b\,\,\vec c\,\,\vec r]}{[\vec a\,\,\vec b\,\,\vec c]},y=\frac{[\vec c\,\,\vec a\,\,\vec r]}{[\vec a\,\,\vec b\,\,\vec c]},z=\frac{[\vec a\,\,\vec b\,\,\vec r]}{[\vec a\,\,\vec b\,\,\vec c]}$

Substituting the values of x, y, z in (i), we get

$[\vec b\,\,\vec c\,\,\vec r]\vec a+[\vec c\,\,\vec a\,\,\vec r]\vec b+[\vec a\,\,\vec b\,\,\vec r]\vec c=[\vec a\,\,\vec b\,\,\vec c]\vec r$