Answer :

Given, Δ =

Applying Elementary transformations, we get


R1 R1 + R2 + R3 we have,



Δ = 2 (a + b + c)


Now applying C2 C2 – C1 and C3 C3 – C1


Δ = 2 (a + b + c)


Expanding along R1


Δ = 2 (a + b + c) [1 {(b – c)(c – b) – (b – a)(c – a)} – 0 + 0]


Δ = 2 (a + b + c) [- b2 – c2 + 2bc – bc + ba + ac – a2]


Δ = 2 (a + b + c) [ab + bc + ca – a2 – b2 – c2]


Given that Δ = 0


2 (a + b + c) [ab + bc + ca – a2 – b2 – c2] = 0


Either a + b + c = 0, or ab + bc + ca – a2 – b2 – c2 = 0


Now


ab + bc + ca – a2 – b2 – c2 = 0


Multiplying both sides by -2


- 2ab – 2bc – 2ca + 2a2 + 2b2 + 2c2 = 0


a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ca + a2 = 0


(a – b)2 + (b – c)2 + (c – a)2 = 0


Since, (a – b)2, (b – c)2, (c – a)2 are non-negative


(a – b)2 = (b – c)2 = (c – a)2


(a – b) = (b – c) = (c – a)


a = b = c


Hence, if ∆ = 0, then either a + b + c = 0 or a = b = c


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