If (2α+1,α2−3α, α−12)\left(2\alpha+1, \alpha^{2}-3\alpha, \;\dfrac{\alpha-1}{2}\right)(2α+1,α2−3α,2α−1) is the image of (α, 2 α, 1) in the line x−23= y−12= z1\;\dfrac{x-2}{3}=\;\dfrac{y-1}{2}=\;\dfrac{z}{1}3x−2=2y−1=1z, then the possible value(s) of α is(are) :
Only 3
Only 3 and -1
Only 3, 143, \;\dfrac{1}{4}3,41 and -1
Only 3 and 14\;\dfrac{1}{4}41