[tex]\displaystyle\it\\\frac{a}{2b+3c}=\frac{2b}{a+3c}=\frac{3c}{a+2b},~atunci~valoarea~expresiei~\\(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)~este~?\\--------------------\\notam~a=x,~2b=y~iar~3c=z.\\\frac{x}{y+z}=\frac{y}{x+z}=\frac{z}{x+y}=m.\\\frac{x}{y+z}=m \implies \frac{x}{y+z}+1=m+1 \implies \frac{y+z}{x+y+z}=\frac{1}{m+1}.\\\frac{y}{x+z}=m \implies \frac{y}{x+z}+1=m+1 \implies \frac{x+z}{x+y+z}=\frac{1}{m+1}.\\[/tex]
[tex]\displaystyle\it\\\frac{z}{x+y}=m \implies \frac{z}{x+y}+1=m+1 \implies \frac{x+y}{x+y+z}=\frac{1}{m+1}.\\adunam~cele~trei~egalitati:\\\frac{2(x+y+z)}{x+y+z}=\frac{3}{m+1} \implies m+1=\frac{3}{2} \implies \boxed{\it m=\frac{1}{2}}.\\-------------------------\\\frac{a}{2b+3c}=\frac{1}{2} \implies 2a=2b+3c~(1).\\\frac{2b}{a+3c}=\frac{1}{2} \implies 4b=a+3c \implies 2b=\frac{a+3c}{2}~(2).\\\frac{3c}{a+2b}=\frac{1}{2} \implies 6c=a+2b \implies 3c=\frac{a+2b}{3}.\\[/tex]
[tex]\displaystyle\it\\inlocuind~in~(1)~rezulta~ca~\boxed{\it c=\frac{1}{3}a}.\\inlouind~ce~am~obtinut~mai~sus~in~(2) \implies \boxed{\it b=\frac{1}{2}a}.[/tex]
[tex]\displaystyle\it\\inlocuim~ce~am~obtinut.\\(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)=\frac{11}{6}a\cdot\frac{6}{a}=\boxed{\it 11}.[/tex]
