Given:
The system of equations,
2x+3y−z=5
x+αy+3z=−4
3x−y+βz=7 have many solutions,
So, △=△1=△2=△3
Now, finding △2=∣2135−47−13β∣=0
⇒2(−4β−21)−5(β−9)−1(7+12)=0
⇒13β=−16
Now, finding △3=∣2133α−15−47∣=0
⇒14α−8−57−5−15α=0
⇒α=−70
Hence, the value of 13αβ=70×16=1120