# If a, b and

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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