+118677 \(a(a+2b)+b(b+2c)+c(c+2a)=\frac{104}{3}+\frac{7}{9}+-7\\ a^2+2ab+b^2+2bc+c^2+2ac=\frac{256}{9}\\ a^2+2ab+2ac+b^2+2bc+c^2=\frac{256}{9}\\ a(a+2b+2c)+b(b+2c)+c^2=\frac{256}{9}\\ a(a+b+c)+a(b+c)+b(b+c)+bc+c^2=\frac{256}{9}\\ a(a+b+c)+(a+b)(b+c)+bc+c^2=\frac{256}{9}\\ a(a+b+c)+(a+b)(b+c)+c(b+c)=\frac{256}{9}\\ a(a+b+c)+(a+b+c)(b+c)=\frac{256}{9}\\ (a+b+c)(a+b+c)=\frac{256}{9}\\ |a+b+c|=\frac{16}{3}\)
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+99 Thanks Melody I appreciate it and it makes way more sense now
