Correct Option: A

Given that :- a² + b² + c² = 2 (a – b – c) – 3
⇒ a² + b² + c² - 2 (a – b – c) + 3 = 0
⇒ a² + b² + c² - 2a + 2b + 2c + 3 = 0
⇒ ( a² + 1 - 2a ) + ( b² + 1 + 2b ) + ( c² + 1 + 2c ) = 0
⇒ ( a - 1 )² + ( b + 1 )² + ( c + 1 )² = 0
This is possible when ( a - 1 )² = 0, ( b + 1 )² = 0, and ( c + 1 )² = 0
⇒ a = 1, b = –1, c = –1
Thus, 2a – 3b + 4c = 2 (1) – 3 (–1) + 4 (–1)
2a – 3b + 4c = 2 + 3 – 4 = 1.