Two lines L1:x=5,y3−α=z−2 and L2:x=α,y−1=z2−α are coplanar. Then α can take value(s)

L1:x−50=y3−α=z−2L2:x−a0=y−1=z2−αAs L1,L2 are coplanar, therefore 5−α0003−α−20−12−α=0⇒(5−α)6−5α+α2−2=0⇒(5−α)(α−1)(α−4)=0⇒α=1,4,5