Which one among the following is correct if x, y, z are eliminated from, (bx/y+z = a, cy/z+x = b, az/x+y = c)?

We are given the following set of equations:

$$\frac{bx}{y+z}=a\text{(Equation 1)}$$

$$\frac{cy}{z+x}=b\text{(Equation 2)}$$

$$\frac{az}{x+y}=c\text{(Equation 3)}$$

Let us rewrite these equations in terms of ratios of the variables $x$, $y$, and $z$.
From Equation 1, we can write:
$$bx=a(y+z)\Rightarrow \frac{x}{y+z}=\frac{a}{b}$$
Adding 1 to both sides (componendo):
$$\frac{x}{y+z}+1=\frac{a}{b}+1\Rightarrow \frac{x+y+z}{y+z}=\frac{a+b}{b}$$
Taking the reciprocal:
$$\frac{y+z}{x+y+z}=\frac{b}{a+b}\text{(Equation 4)}$$

Similarly, using Equation 2, we have:
$$cy=b(z+x)\Rightarrow \frac{y}{z+x}=\frac{b}{c}$$
Adding 1 to both sides:
$$\frac{x+y+z}{z+x}=\frac{b+c}{c}$$
Taking the reciprocal:
$$\frac{z+x}{x+y+z}=\frac{c}{b+c}\text{(Equation 5)}$$

Similarly, using Equation 3, we have:
$$az=c(x+y)\Rightarrow \frac{z}{x+y}=\frac{c}{a}$$
Adding 1 to both sides:
$$\frac{x+y+z}{x+y}=\frac{c+a}{a}$$
Taking the reciprocal:
$$\frac{x+y}{x+y+z}=\frac{a}{c+a}\text{(Equation 6)}$$

Now, let us add Equation 4, Equation 5, and Equation 6:
$$\frac{y+z}{x+y+z}+\frac{z+x}{x+y+z}+\frac{x+y}{x+y+z}=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
Simplifying the left-hand side:
$$\frac{2(x+y+z)}{x+y+z}=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
$$2=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$

Subtracting 1 from both sides of the Equation:
$$1=\left(\frac{b}{a+b}-1\right)+\frac{c}{b+c}+\frac{a}{c+a}$$
$$1=\frac{-a}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
Let us rearrange the terms to simplify calculations:
$$1-\frac{c}{b+c}=\frac{a}{c+a}-\frac{a}{a+b}$$
$$\frac{b}{b+c}=a\left(\frac{1}{c+a}-\frac{1}{a+b}\right)$$
$$\frac{b}{b+c}=a\left(\frac{(a+b)-(c+a)}{(c+a)(a+b)}\right)$$
$$\frac{b}{b+c}=\frac{a(b-c)}{(c+a)(a+b)}$$

Cross-multiplying the terms:
$$b(c+a)(a+b)=a(b-c)(b+c)$$
$$b(ac+bc+a^2+ab)=a(b^2-c^2)$$
$$abc+b^{2}c+a^{2}b+a{b}^2=a{b}^2-a{c}^2$$

Subtracting $a{b}^2$ from both sides, we get:
$$a^{2}b+b^{2}c+abc=-c^{2}a$$
Rearranging all the terms to one side:
$$a^{2}b+b^{2}c+c^{2}a+abc=0$$

This matches the relationship in the correct answer.