If a^{2}+b^{2}-c^{2}=2(a-b-c)-3, then the val | vedclass

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(A) Given the equation: $a^{2}+b^{2}-c^{2}=2(a-b-c)-3$Rearranging the terms: $a^{2}+b^{2}-c^{2}-2a+2b+2c+3=0$Note: There is a sign correction required

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$2$

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(A) Given the equation: $a^{2}+b^{2}-c^{2}=2(a-b-c)-3$Rearranging the terms: $a^{2}+b^{2}-c^{2}-2a+2b+2c+3=0$Note: There is a sign correction required in the original equation to make it solvable as a sum of squares. Assuming the equation is $a^{2}+b^{2}+c^{2}=2(a-b-c)-3$:$a^{2}-2a+1 + b^{2}+2b+1 + c^{2}+2c+1 = 0$$(a-1)^{2} + (b+1)^{2} + (c+1)^{2} = 0$Since the sum of squares is zero,each term must be zero:$a-1=0 \Rightarrow a=1$$b+1=0 \Rightarrow b=-1$$c+1=0 \Rightarrow c=-1$Now,calculate $4a-3b+5c$:$4(1) - 3(-1) + 5(-1) = 4 + 3 - 5 = 2$

If $a^{2}+b^{2}-c^{2}=2(a-b-c)-3,$ then the value of $4a-3b+5c$ is

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