Question

if a2+b2+c2+3=2(a-b-c), then 3a-2ab+c?

Answer 1

${a}^{2} + {b}^{2} + {c}^{2} + 3 = 2(a-b-c ) \\ {a}^{2} + {b}^{2} + {c}^{2} + 3 = 2a - 2b - 2c \\( {a}^{2} -2a + 1) +( {b}^{2} +2b + 1) + ( {c}^{2} + 2c + 1) = 0 \\( a - 1) ^{2} + (b - 1)^{2}+ (c - 1) ^{2} = 0$
then,(a-1)^2=0
a-1=0
a=1

(b+1)^2=0
b=-1

(c+1)^2=0
c=-1

$then \\ 3a - 2ab + c = 3 \times 1 - 2 \times 1 \times ( - 1) + ( - 1) = 0$