Correct Answer - 3
`x+y+z=1`
`x+2y +4z=p`
`x+4y+10z=p^(2)`
`Delta = |{:(1,,1,,1),(1,,2,,4),(1,,4,,10):}|`
`R_(1) to R_(1)-R_(2),R_(2),R_(2) to R_(2)-R_(3)`
`= |{:(0,,-1,,-3),(-,,-2,,-6),(1,,4,,10):}|=0`
Since `Delta =0` solutions is not unique
The system will have infinite solutions if `Delta_(1)=0 ,Delta_(2) =0 Delta_(3)=0`
`Delta_(1)= |{:(1,,1,,1),(p,,2,,4),(p^(2),,4,,10):}|=0`
`C_(3) to C_(3)-C_(2)`
`Delta_(1)= |{:(1,,1,,0),(p,,2,,2),(p^(2),,4,,6):}|=0`
`" or " 1(12 -8) -1 (6p -2p^(2)) =0`
`" or " 4-6p +2p^(2) =0`
`" or " 2(p^(2) -3p+2) =0`
`" or " p^(2) -3p +2=0`
`rArr p=1" or "2`
Also for these values of `p,Delta_(2),Delta_(3)=0`
