As the lines intersect at the x−axis, let point of intersection of lines be P(k,0)
Therefore point P must satisfy the given equation
⇒6k2−24k+β=0
There is only one such k possible for the intersection of the two lines and so the above equation must have repreated roots.
⇒b2−4ac=0
⇒β=24 and k=2
Substitute (2,y) back into the equation
6(2)2−α(2)y−3y2−24(2)+3y+24=0⇒y2+(3−2α)y=0
But we know that y=0 is the only solution to the above equation, so even this equation has repeated roots.
⇒b2−4ac=0⇒α=32
Hence 20α−β=20×32−24=6