We have,

AB=(−2−1)2+(3−9)2=45
AC=(−2−3)2+(3−8)2=50
BC=(3−1)2+(8−9)2=5
Here,
AC2=AB2+BC2
⇒(50)2=(45)2+(5)2
Hence, ∠B=90∘, so circumcentre is the mid-point of hypotenuse i.e., AC.
Circumcenter ≡(21,211)
Mid-point of BC=(2,217)
Equation of line L is
(y−211)=(21−2211−217)(x−21)
⇒y=2x+29
Passing though (0,2α), so
2α=29⇒α=9