Let us factorise

Given, 3a² – b² – c² + 2ab – 2bc + 2ca

⇒ 4a² – a² – b² – c² + 2ab – 2bc + 2ca

⇒ (2a)² – (a² + b² + c² – 2ab + 2bc – 2ca)

⇒ (2a)² – (a – b – c)²

[Since, from the identity III we know that, (x² – y²) = (x – y)(x + y)]

⇒ (2a – (a – b – c))(2a + a + b + c)

⇒ (a + b + c)(3a + b + c)