Andrew's three symmetric numbers

As per the question,we get these equations: $\begin{aligned} a+c-b= & 1 \quad \quad \quad \ldots (1) \\ a+b-c= & 2 \quad \quad \quad \ldots (2) \\ b+c-a= & 3 \quad \quad \quad \ldots (3) \end{aligned}$ Adding $(1)$ and $(2)$ ,we get $\begin{aligned} a+c-b+a+b-c= & 1+2 \\ \implies 2a= & 3 \\ \therefore a= & \dfrac{3}{2} \end{aligned}$ Adding $(2)$ and $(3)$ ,we get $\begin{aligned} a+b-c+b+c-a= & 2+3 \\ \implies 2b= & 5 \\ \therefore b= & \dfrac{5}{2} \end{aligned}$ Adding $(1)$ and $(3)$ ,we get $\begin{aligned} a+c-b+b+c-a= & 1+3\\ \implies 2c= & 4 \\ \therefore c= & 2 \end{aligned}$ Now when we have got the three numbers,we can calculate the required value as $\dfrac{b+c}{a}=\dfrac{\frac{5}{2}+2}{\frac{3}{2}}=\boxed{3}$

Yes as @Kushagra Sahni said :- $\begin{aligned} a + c - b = 1 \cdots (1) \\ a + b - c = 2 \cdots (2) \\ b + c - a = 3 \cdots (3) \end{aligned}$ Now adding all the equations we get, $\begin{aligned} &a + b + c = 6 \\ &b + c = 6 - a \cdots (4) \end{aligned}$ From $(3)$ we get $b + c = 3 + a$

$\therefore$ from $(3)$ and $(4)$ $a = \dfrac{3}{2}$

$\therefore \dfrac{b + c}{a} = \dfrac{6 - a}{a} = \dfrac{9}{3} = 3$