Algebra experts can answer this in a second

$\begin{aligned} (a+b+c)\left(\dfrac { 1 }{ a } +\dfrac { 1 }{ b } +\dfrac { 1 }{ c } \right) &= \dfrac { a }{ b } +\dfrac { c }{ b } +\dfrac { b }{ a } +\dfrac { c }{ a } +\dfrac { b }{ c } +\dfrac { a }{ c }+ 3 \\ \Rightarrow 3 \times 5 &= \dfrac { a }{ b } +\dfrac { c }{ b } +\dfrac { b }{ a } +\dfrac { c }{ a } +\dfrac { b }{ c } +\dfrac { a }{ c }+ 3 \\ \Rightarrow \boxed{12} &=\dfrac { a }{ b } +\dfrac { c }{ b } +\dfrac { b }{ a } +\dfrac { c }{ a } +\dfrac { b }{ c } +\dfrac { a }{ c } \end{aligned}$

$\begin{aligned} \frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}\\=\frac{a+c}{b}+\frac{a+b}{c}+\frac{b+c}{a}\\=\frac{3-b}{b}+\frac{3-c}{c}+\frac{3-a}{a}\\ =\frac{3}{b}+\frac{3}{a}+\frac{3}{c}-3\\ = \frac{3(ab+ac+bc)}{abc}-3 \end{aligned}$ From $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=5$ ,we get: $\begin{aligned}\frac{ab+ac+bc}{abc}=5\\ab+ac+bc=5abc\\ \rightarrow 3(ab+ac+bc)=15abc \end{aligned}$ Hence $\dfrac{3(ab+ac+bc)}{abc}-3=\dfrac{15abc}{abc}-3=15-3=\boxed{12}$

a/b+a/c+b/a+b/c+c/a+c/b=(b+c)/a+(a+c)/b+(a+b)/c=(3-a)/a+(3-b)/b+(3-c)/c=3(1/a+1/b+1/c)-3=3*5-3=12

Its my solution a/b +a/c +b/a +b/c +c/a +c/b =a(1/b+1/c) + b(1/a+1/c) + c(1/a+1/b) =a(5-1/a)+b(5-1/b) +c(5-1/c) =5a-1 +5b-1 +5c-1 =5(a+b+c)-3 =5*3-3 =12

Just multiple the 2 equations