Given:
The system of linear equations
x−2y+z=−4,
2x+αy+3z=5 and
3x−y+βz=3 has infinity many solutions,
So, △=△1=△2=△3
Now, finding,
△2=∣123−45313β∣=0
⇒5β−9+8β−36−9=0
⇒13β=54
And, △3=∣123−2α−1−453∣=0
⇒3α+5+12−30+8+12α=0
⇒15α=5
⇒12α=4
Hence, the value of 12α+13β=4+54=58