Correct Answer - C
the given system is consistent . Therefore
`Delta =|{:(1,,1,,-1),(2,,-1,,-c),(-b,,3b,,-c):}|=0`
`" or " c+bc -6b +b+ 2c+ 3bc=0`
`" or " 3c+4bc-5b=0`
`" or " 3c+ 4bc -5b =0`
`" or " c=(5b)/(4b+3)`
Now `c lt 1`
`rArr (5 b)/(4b+3) lt 1 `
`" or " .(5b)/(4b +3) -1 lt 0`
`rArr b in (-(3)/(4),3)`