Correct Answer - 25
`Delta = |{:(a,,b+3c,,c+4a),(b,,c+3a,,a+4b),(c,,a+3b,,b+4c):}|+2 |{:(b,,b+3c,,c+4a),(c,,c+3a,,a+4b),(a,,a+3b,,b+4c):}|`
Applying `C_(3) to C_(3) -4C_(1) " in " Delta " and " C_(2) to C_(2) -C_(1) " in " Delta'` we get
`Delta_(2)= |{:(a,,b+3c,,c),(b,,c+3a,,a),(c,,a+3b,,b):}|+2xx3 |{:(b,,c,,c+4a),(c,,a,,a+4b),(a,,b,,b+4c):}|`
applying `c_(2) to C_(2)- 3C_(3) " in " Delta " and " C_(3) to C_(3) -C_(2) " in " Delta'` we get
` = |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|+6 |{:(b,,c,,4a),(c,,a,,4b),(a,,b,,4c):}|`
`=|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|+24 |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|`
`=25 |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}|=25Delta_(1)`
`rArr (Delta_(2))/(Delta_(1)) =25`