Question: If the system of linear equations :
$x+a y+z=3$
$x+2 y+2 z=6$
$x+5 y+3 z=b$
has no solution, then :-
$\mathrm{a}=-1, \mathrm{~b}=9$
$a \neq-1, b=9$
$\mathrm{a}=1, \mathrm{~b} \neq 9$
$\mathrm{a}=-1, \mathrm{~b} \neq 9$
Correct Option: 1,
Solution:
Solution Not Required
