‌‌ (i) ‌‌‌‌‌‌‌‌‌‌‌‌‌‌ (ii) ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ (iii) ‌ From (i) and (ii) we get ‌|4a+9|=15 ‌⇒4a+9=15,−15 ‌⇒‌‌4a=6,−24⇒a=‌
3
2
,−6 From (i) and (iii), we get 3α+4β−3=±15 Case (I) When 3α+4β−3=15‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) Also, 4α−3β+1=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) (given) On solving Eqs. (i) and (ii) we get α=2β=3 Case (II) When 3α+4β−3=−15 ⇒‌‌3α+4β=−12‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii) ‌ and ‌4α−3β+1=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv) On solving Eqs. (iii) and (iv) we get, α=‌
−8
5
,β=‌
−9
5
∴ Sum of all possible values of a,α,β=‌
3
2
+2+3=‌
3
2
+5=‌
13
2
‌‌‌⋅⋅⋅⋅⋅⋅⋅(v) ‌ or ‌−6−‌
8
5
−‌
9
5
=‌
−30−17
5
=‌
−47
5
‌‌‌⋅⋅⋅⋅⋅⋅⋅(vi) On adding Eqs. (v) and (vi) ∴ Total sum =‌