$$\text{If}\quad a= \frac{x}{(2x+y+z)} = \frac{y}{(x+2y+z)} =\frac{z}{(x+y+2z)} ,$$ then find the value of a.

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Correct option is B

Given:

​$$a= \frac{x}{(2x+y+z)} = \frac{y}{(x+2y+z)} =\frac{z}{(x+y+2z)}$$ 

Solution:

Let us define the equations as follows:

​$$a = \frac{x}{2x + y + z}, \quad a = \frac{y}{x + 2y + z}, \quad a = \frac{z}{x + y + 2z}$$ 

Express $x$, $y$, and $z$ in terms of each other

$$a = \frac{x}{2x + y + z} \quad \Rightarrow \quad x = a(2x + y + z)$$ 

$$x = 2ax + ay + az$$ 

​$$x - 2ax = ay + az \quad \Rightarrow \quad x(1 - 2a) = ay + az$$ 

​$$x = \frac{ay + az}{1 - 2a} \quad \text{(Equation 1)}$$ 

$$a = \frac{y}{x + 2y + z} \quad \Rightarrow \quad y = a(x + 2y + z)$$ 

$$y = ax + 2ay + az$$ 

$$y - 2ay = ax + az \quad \Rightarrow \quad y(1 - 2a) = ax + az$$ 

$$y = \frac{ax + az}{1 - 2a} \quad \text{(Equation 2)}$$ 

$$a = \frac{z}{x + y + 2z} \quad \Rightarrow \quad z = a(x + y + 2z)$$ 

$$z = ax + ay + 2az$$ 

$$z - 2az = ax + ay \quad \Rightarrow \quad z(1 - 2a) = ax + ay$$ 

$$z = \frac{ax + ay}{1 - 2a} \quad \text{(Equation 3)}$$ 

​$$\text{Since the equations are symmetric in } x, y, \text{ and } z, \text{ assume } x = y = z = k.$$  

​$$\text{Substitute } x = y = z = k \text{ into the first equation:}$$ 

$$a = \frac{k}{2k + k + k} = \frac{k}{4k} = \frac{1}{4}$$ 

​Thus, the value of $a$ is  $$\frac{1}{4}$$.​

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