Test Your Basic Concept....................

$\small (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=6 \times \frac{3}{2}=9$ $\iff \frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}=9$ $\iff \large (\color{#3D99F6}{\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}})+(\frac{a}{a}+\frac{b}{b}+\frac{c}{c})=9$ $\iff \Large (\color{#3D99F6}{\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}})+1+1+1=9$ $\iff \LARGE \color{#3D99F6}{\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}}=9-1-1-1=\Huge \color{#20A900}{\boxed{\color{#3D99F6}{6}}}$

since a +b +c = 6 now a/b+a/c+b/a+b/c + c/b+ c/a = a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a) = a(3/2-1/a)+b(3/2-1/b)+c(3/2-1/c) =3/2(a+b+c)-3 =3/2(9) - 3 = 6

Good problem but you can guess $a = 2, b = 2, c = 2$

A proper solution would be to add $\frac { a }{ a } ,\frac { b }{ b } ,\frac { c }{ c }$ . Then, the desired value will increase by 3 and can be obtained by multiplying $(a+b+c)(\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } )=9\\ 9-3=6$