The system of linear equations
x+y+z=62x+5y+az=36x+2y+3z=b has [2026]
(3)
If D=|11125a123|=0⇒a=8
If D1=|611365ab23|=0⇒ab-5b-12a+54=0
If D2=|161236a1b3|=0⇒ab-6a-2b-36=0
If D3=|116253612b|=0⇒b=14
For a=8 and b=14⇒D1,D2 are also zero
For a=8 and b=14⇒D=D1=D2=D3=0
⇒infinitely many solutions.