Using properties of determinants, prove that
.(CBSE 2018)

we get,
= 
Taking (a + b + c) common we get,
= 
R2→ R2 – R1 and R3→ R3 – R1 we get,
= 
Taking 2 common from R2 and R3
= 
= 4(a + b + c)((a + 2b)(a + 2c)–(a–b)(a–c))
= 4(a + b + c)(a2 + 2ac + 2ac + 4bc–a2 + ac + ab–bc)
= 4(a + b + c)3(ab + bc + ca) = 12(a + b + c)(ab + bc + ca)
L.H.S = R.H.S
Hence, proved
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![[ ccc -4&4&4 -7&1&3 5&-3&-1 ] [ ccc 1&-1&1 1&-2&-2 2&1&3 ]](https://static.philoid.co/ncertusercontent/solutions/?domain=gF&l=PROJ14073/1521536408693719.png)