show that `|{:(3a,,-a+b,,-a+c),(-b+a ,,3b,,-b+c),( -c+a,, -c+b,,3c):}|`
`= 3(a+b+c) (ab+ bc+ca)`

Explanation

`=|{:(3a,-a+b,-a+c),(-b+a,3b,-b+c),(-c+a,-c+b,3c):}|=|{:(a+b+c,-a+b,-a+c),(a+b+c,3b,-a+c),(a+b+c,-c+b,3c):}|`
`(C_(1)toC_(1)+C_(2)+C_(3))`
`=(a+b+c)|{:(1,-a+b,-a+c),(1,3b,-b+c),(1,-c+b,3c):}|`
`=(a+b+c)|{:(1,-a+b,-a+c),(0,a+2b,a-b),(0,a-c,a+2c):}|`
`(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))`
`=(a+b+c).1|{:(a+2b,a-b),(a-c,a+2c):}|`
(Expanding along ` C_(1)`)
=(a+b+c)[(a+2b)(a+2b)(-a-b)(a-c)]
`=(a+b+c)(a^(2)+2ab+2ab+4bc-a^(2)+ac+ab-bc)`
=(a+b+c)(3ab+3bc+3ac)
=3(a+b+c)(ab+bc+ca)
=R.H.S.