For some non-zero real values of $$a, b$$ and $$c$$, it is given that $$\mid \frac{c}{a} \mid = 4, \mid \frac{a}{b} \mid = \frac{1}{3}$$ and $$\frac{b}{c} = -\frac{3}{4}$$. If $$ac > 0$$, then $$\left(\frac{b + c}{a}\right)$$

Since the product $$ac$$ is greater than zero, we get that either both $$a$$ and $$c$$ are positive, or both are negative. $$b$$ will have the opposite sign as $$a$$ and $$c$$ because the fraction $$\frac{b}{c}$$ is negative. This gives us the following cases.

1. Case 1: Both $$a$$ and $$c$$ are positive, and $$b$$ is negative.
   
   We will conclude $$\frac{c}{a}=4$$, $$\frac{a}{b}=\frac{-1}{3}$$, and $$\frac{b}{c}=\frac{-3}{4}$$
   Therefore $$\frac{b}{a}=-3$$ and we get $$\frac{b+c}{a} = \frac{c}{a}+\frac{b}{a} = 4-3=1$$
2. Case 2: Both $$a$$ and $$c$$ are negative, and $$b$$ is positive
   
   The calculations will be the same.

Therefore, the correct answer is $$1$$.