If the system of equations x+y+z=5, x+2y+3z=9, x+3y+αz=β has infinitely many solutions, then β-2α equals:
(2)
x+y+z=5, x+2y+3z=9, x+3y+αz=β
D=|11112313α|=0
⇒(2α-9)+(3-α)+(3-2)=0
⇒α=5
Now, D3=|11512913β|=0
⇒2β-27+9-β+5(3-2)=0
⇒β=13
⇒at α=5, β=13