已知 $△ABC$为等边三角形， $AB=2$，设点 $P,Q$满足 $\stackrel{\rightharpoonup }{AP}=\lambda \stackrel{\rightharpoonup }{AB}$， $\stackrel{\rightharpoonup }{AQ}=(1-\lambda )\stackrel{\rightharpoonup }{AC}$， $\lambda \in R$，若 $\stackrel{\rightharpoonup }{BQ}·\stackrel{\rightharpoonup }{CP}=\frac{3}{2}$，则 $\lambda$=

A．	$\frac{1}{2}$	B．	$\frac{1\pm \sqrt{2}}{2}$
C．	$\frac{1\pm \sqrt{10}}{2}$	D．	$\frac{-3\pm 2\sqrt{2}}{2}$