proof for $ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$ this formula just pop up in textbook I'm reading without any explanation
$ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$
I did some "vector arithmetic" using the determinant method
but I'm not getting the answer to agreed with above formula.
I'm wondering if anyone saw this formula before and know the proof of it?
the final result that i get is
$b_{1}(a_{3}c_{3}+a_{2}c_{2})-a_{1}(b_{2}c_{2}+b_{3}c_{3})i$
$b_{2}(a_{3}c_{3}+a_{1}c_{1})-a_{2}(b_{3}c_{3}+b_{1}c_{1})j$
$b_{3}(a_{2}c_{2}+a_{1}c_{1})-a_{3}(b_{2}c_{2}+b_{1}c_{1})k$
But I failed to see any correlation for $(\vec{A}\cdot\vec{C})$ and $(\vec{B}\cdot\vec{C})$ part...

vector $ \vec A\times \vec B$ is perpendicular to the plane containing $\vec A $ and $\vec B$.Now, $(\vec A\times \vec B)\times \vec C$ is perpendicular to plane containing vectors $\vec C$ and $\vec A\times \vec B$, thus $(\vec A\times \vec B)\times \vec C$ is in the plane containing $\vec A$ and $ \vec B$ and hence $(\vec A\times \vec B)\times \vec C$ is a linear combination of vectors $\vec A$ and $ \vec B\implies (\vec A\times \vec B)\times \vec C=\alpha \vec A + \beta \vec B$. Take dot product with $\vec C$ both sides gives $0$ on the L.H.S(as $(\vec A\times \vec B)\times \vec C$ is perpendicular to $\vec C$), hence $0=\alpha (\vec A. \vec C)+\beta(\vec B.\vec C)\implies \frac{\beta}{\vec A. \vec C}=\frac{-\alpha}{\vec B. \vec C}=\lambda \implies \alpha=-\lambda(\vec B. \vec C)$ and $\beta=\lambda(\vec A. \vec C) \implies (\vec A\times \vec B)\times \vec C=\lambda((\vec A. \vec C)\vec B-(\vec B. \vec C)\vec A)$. Here, $\lambda$ is independent of the magnitudes of vectors as if magnitude of any vector is multiplied by any scalar, that appears explicitly on both sides of the equation.Thus , putting unit vectors $\vec i,\vec j,\vec i$ in the equation gives $\vec j=\lambda(\vec j)\implies \lambda=1$ and hence $(\vec A\times \vec B)\times \vec C=((\vec A. \vec C)\vec B-(\vec B. \vec C)\vec A)$   
