Proof of $BAC-CAB$ identity missing step I'm stuck on one step of the proof for the identity: $$ \vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$$
So far, the proof follows as:
We know that $\vec{B}\times\vec{C}$ gives a vector perpendicular to both $\vec{B}$ & $\vec{C}$, and that $ \vec{A}\times(\vec{B}\times\vec{C})$ gives a vector perpendicular to both $\vec{A}$ & $(\vec{B}\times\vec{C})$. Therefore, the vector $\vec{A}\times(\vec{B}\times\vec{C})$ must lie in the plane containing both $\vec{B}$ & $\vec{C}$.
Provided $\vec{B}$ & $\vec{C}$ are not parallel (if they were, $\vec{A}\times(\vec{B}\times\vec{C}) = 0$ regardless), vectors $\vec{B}$ & $\vec{C}$ span the 2D plane containing them both.
Therefore, we can  express any vector in the plane as a linear combination of both $\vec{B}$ and $\vec{C}$ and so we can write: $$\vec{A}\times(\vec{B}\times\vec{C}) = \alpha\vec{B} + \beta\vec{C} \tag{1}$$
Taking the scalar product of both sides with $\vec{A}$: $$\vec{A} \cdot (\vec{A}\times(\vec{B}\times\vec{C})) = \vec{A} \cdot (\alpha\vec{B} + \beta\vec{C}) = 0$$ So, $$\alpha(\vec{A} \cdot \vec{B}) + \beta(\vec{A} \cdot\vec{C}) = 0$$
Now writing, $$\lambda = \frac{\alpha}{\vec{A} \cdot\vec{C}} = -\frac{\beta}{\vec{A} \cdot \vec{B}}$$
and substituting $\alpha$ and $\beta$ back into (1) we get:
$$ \vec{A}\times(\vec{B}\times\vec{C}) = \lambda(\vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B}))$$
I am able to show $\lambda = 1$ with particular choices of unit vectors for $\vec{A}, \vec{B}, \vec{C}$ but I am unable to prove that $\lambda$ is independent of the magnitude of vectors (i.e. $\lambda = 1$ for all choices of $\vec{A}, \vec{B}, \vec{C}$). This is the step that I am struggling with. Any suggestions?

Observe that both sides of the equation to be proved are linear in each of the three variables (when the other two are fixed).
Based on this, it's enough to prove the statement for all possible combinations of the standard basis, say.
