Mark the tick against the correct answer in the following:

To find: Value of

We have,

Applying

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Taking (a - b) common

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Applying

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Taking (c-a) common

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Expanding along R₁

= (b – a)(c – a)[0 – 1(1 – (a – b)) + (b + c)(0)]

= (b – a)(c – a)(-1 + a – b)

= (b – a)(c – a)(a – b – 1)

= (b – a)(ac – bc – c – a² + ab + a)

= (abc – b²c – bc – a²b + ab² + ab – a²c + abc + ac + a³ + a²b + a²)

= 4abc