We are given that
a^2 + b^2 + c^2 = 2a - 2b - 2
=> a^2 + b^2 + c^2 - 2a + 2b + 2 = 0
=> a^2 - 2a +1 + b^2 + 2b + 1 + c^2 = 0
=> (a - 1)^2 + (b + 1)^2 + c^2 = 0
(a - 1)^2, (b + 1)^2, and c^2 are all squares, which means all are >/= 0. As the minimum value of each of them individually is 0, their sum could only be 0 when all of them become equal to 0.
(a - 1)^2 is equal to 0 at a = 1
(b + 1)^2 is equal to 0 at b = -1
c^2 is equal to 0 at c = 0
Hence, a = 1, b = -1, and c = 0.
Which means 3a - 2b + c = 3(1) - 2(-1) + 0 = 5
a^2 + b^2 + c^2 = 2a - 2b - 2
=> a^2 + b^2 + c^2 - 2a + 2b + 2 = 0
=> a^2 - 2a +1 + b^2 + 2b + 1 + c^2 = 0
=> (a - 1)^2 + (b + 1)^2 + c^2 = 0
(a - 1)^2, (b + 1)^2, and c^2 are all squares, which means all are >/= 0. As the minimum value of each of them individually is 0, their sum could only be 0 when all of them become equal to 0.
(a - 1)^2 is equal to 0 at a = 1
(b + 1)^2 is equal to 0 at b = -1
c^2 is equal to 0 at c = 0
Hence, a = 1, b = -1, and c = 0.
Which means 3a - 2b + c = 3(1) - 2(-1) + 0 = 5
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