If a^2 +b^2 + 4c^2 = 2(a+b-2c)-3,and a,b,c are real, then the value of (a^2+b^2+c^2)is?

##2 1/4##

##a^2+b^2+4c^2=2(a+b-2c)-3## Rearrange and set = 0: ##a^2+b^2+4c^2-2(a+b-2c)+3=0## Expand: ##a^2+b^2+4c^2-2a-2b+4c+3=0## Organize like terms: ##a^2-2a+b^2-2b+4c^2+4c+3=0## Rearrange for perfect squares form: ##a^2-2a+1+b^2-2b+1+4c^2+4c+1=0## Factor: ##(a-1)^2+(b-1)^2+(2c+1)^2=0## Set all three = 0 to satisfy identity: ##(a-1)^2=0,(b-1)^2=0,(2c+1)^2=0## ##a=1, b=1, c=-1/2## Therefore: ##a^2+b^2+c^2 = 1+1+1/4=2.25##