Quadratic equation?

i) Instead of α, let me take the one common root as u.

ii) So substituting in first equation, 3u² + au + b = 0
and from 2nd, 3u² + bu + a = 0

Subtracting these two, (a - b)u + (b - a) = 0
==> (a - b)u = a - b; so u = 1
Thus the value of α = 1

iii) Taking the common root as = 1,
1st equation is: 3 + a + b = 0
and 2nd equation is: 3 + b + a = 0
Adding both and simplifying, a + b = -3 ------------ (1)

iv) x² + hx + k = 0, has roots a & b
Applying properties of roots, a + b = -h ---------- (2) and ab = k -------- (3)

From (1) & (2): h = 3

So the equation is: x² + 3x + k = 0, which two distinct real and rational roots a & b.

For the two roots to be distinct, real and rational,
Its discriminant has to be > 0 and it must be a perfect square.

So here the discriminant is: 9 - 4k > 0
==>-4k > -9
k < 9/4

As k is a positive integer, k = either 1 or 2
When k = 1, discriminant is = 9 - 4 = 5, which is not a perfect square. So this value is rejected.
When k = 2, discriminant is = 9 - 8 = 1, which is a perfect square.
So k = 2

Thus α = 1; h = 3 and k = 2.

NOTE: It appears you have asked repeatedly this question for three times and this is the first solution you are receiving. Taking liberty may I request you as - "Valuing the time and efforts put up by the solvers, kindly acknowledge the solution if you are satisfied with the same, which need not be awarding best solution, but can be a simple thanks.