If (a – 1)2 + (b + 2)2 + (c + 1)2 = 0, then the value

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(a – 1)² + (b + 2)² + (c + 1)² = 0
⇒ a – 1 = 0 ⇒ a = 1;
b + 2 = 0 ⇒ b = –2
c + 1 = 0 ⇒ c = –1
∴ 2a – 3b + 7c
= 2 – 3 (– 2) + 7 (–1)
= 2 + 6 – 7 = 1