Correct Answer: $1$

Solution : Given: $\frac{1}{2+2 p} + \frac{1}{2+2 q} + \frac{1}{2+2 r}$, where $p = \frac{x}{y+z}$, $q=\frac{y}{z+x}$ and $r=\frac{z}{x+y}$
Substitute the values of p, q, and r in the above expression, and we get,
$\frac{1}{2+2\times \frac{x}{y+z}} + \frac{1}{2+2\times \frac{y}{z+x}} + \frac{1}{2+2\times \frac{z}{x+y}}$.
Now, the first term can be written as.
$\frac{1}{2+2\times \frac{x}{y+z}} = \frac{1}{2+\frac{2x}{y+z}}=\frac{y+z}{2y+2z+2x}$
Similarly, the second term can be written as $\frac{1}{2+2\times \frac{y}{z+x}}=\frac{1}{2+\frac{2y}{z+x}}=\frac{z+x}{2z+2x+2y}$
The third term can be written as $\frac{1}{2+2\times \frac{z}{x+y}} = \frac{1}{2+\frac{2z}{x+y}} = \frac{x+y}{2x+2y+2z}$
So, $\frac{y+z}{2y+2z+2x} + \frac {z+x}{2z+2x+2y} + \frac{x+y}{2x+2y+2z}=\frac{2\times(x+y+z)}{2\times(x+y+z)} = 1$
Hence, the correct answer is $1$.