formula: $\hat{b}(\hat{a}\cdot\hat{c})-\hat{c}(\hat{a}\cdot\hat{b})=\hat{a}\times(\hat{b}\times\hat{c})$
$\hat{a}\times(\hat{b}\times\hat{c})$ is on the $\hat{b}$, $\hat{c}$ plane, so:
$\hat{b}r+\hat{c}s=\hat{a}\times(\hat{b}\times\hat{c})$
want to proof:
$r=(\hat{a}\cdot\hat{c})$
$s=-(\hat{a}\cdot\hat{b})$
$\hat{a}\cdot$ both sides:
$\hat{a}\cdot(\hat{b}r+\hat{c}s)=\hat{a}\cdot[\hat{a}\times(\hat{b}\times\hat{c})]$
$(\hat{a}\cdot\hat{b})r+(\hat{a}\cdot\hat{c})s=0$
It seems needing another condition to distinguish $\hat{a}\times(\hat{b}\times\hat{c})$ and $(\hat{b}\times\hat{c})\times\hat{a}$