Taking random values will not do

the above three equation gives $xyz+1=\frac{yz}{5}= \frac{-xz}{15}= \frac{xy}{3}$

now from above equation $3z=5x,3y=-x,-z=5y$

let's see what we have to find, it is $\frac{z-y}{z-x}$ so if we convert $x,y$ in the form of $z$ than it will be quite helpful,

now $x= \frac{3}{5}z$

$y= \frac{-z}{5}$

So at last $\frac{z-y}{z-x}= \frac{z(1+\frac{1}{5})}{z(1-\frac{3}{5})}=3$

so the correct answer is $\boxed{3}$

$\dfrac{xyz+1}{yz}=\dfrac{1}{5}, \dfrac{xyz+1}{xz}=\dfrac{-1}{15}, \dfrac{xyz+1}{xy}=\dfrac{1}{3}$ $\Rightarrow \dfrac{yz}{5}=\dfrac{-xz}{15}=\dfrac{xy}{3}$ $\Rightarrow \dfrac{x}{y}=-3 ~and~ \dfrac{z}{y}=-5$ $\Rightarrow \dfrac{z-y}{z-x}=\dfrac{-5y-y}{-5y+3y}=\Large\color{forestgreen}{3}$

i have done the sum algebrically correct.But logically Idon't agree with the second equation, because x,y,z are real numbers, the equivalent value should not be negative.

By subtracting equations:

(3)-(2)

z+ $\frac {1}{xy}$ -y- $\frac {1}{xz}$ = $\frac {1}{3}$ - $\frac {-1}{15}$

z-y+ $\frac {z-y}{xyz}$ = $\frac {6}{15}$

By taking (z-y) as a common factor:

(z-y)(1+ $\frac {1}{xyz})$ )= $\frac {6}{15}$ ...*

Then we subtract (3)-(1):

z+ $\frac {1}{xy}$ -x- $\frac {1}{yz}$ = $\frac {1}{3}$ - $\frac {1}{5}$

z-x+ $\frac {z-x}{xyz}$ = $\frac {2}{15}$

By taking (z-x) as a common factor:

(z-x)(1+ $\frac {1}{xyz})$ )= $\frac {2}{15}$ ...**

Then:

$\frac {*}{**}$

$\frac {(z-y)}{(z-x)}$ =3