let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$ and : $\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$then : $x+y+z =?$ let $x ,y ,z \in \mathbb{N} ,x \leq y\leq z$
and :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}$$
then :
$$x+y+z =?$$
my try :
$$\frac{100}{336}=\frac{1}{x}+\frac{1}{xy}+\frac{1}{xyz}=\frac{xy+z+1}{xyz}=\frac{25}{84}$$
now ?

Let $x,y,z\in\mathbb{N}$ with $x<y<z$. First, you can rewrite the equation as
$$ \dfrac{100}{336}=\dfrac{25\lambda}{84\lambda}=\dfrac{yz+z+1}{xyz}. $$
for a positive integer $\lambda$. Since we know that $xyz$ and that $yz+z+1$ are  integers, we can hope to find a solution for the system of equations:
$$\begin{cases}84\lambda=xyz\\
24\lambda=y(z+1)\end{cases}.$$
We then have $y=\tfrac{24\lambda}{z+1}$, which translates to 
$$ 84=\dfrac{24xz}{z+1}\qquad\Rightarrow\qquad7(z+1)=2xz.$$
Thus, $7=(2x-7)z$, which suggests the solution $x=4$ and $z=7$. Inputting this in the first equation of the system, we have $3\lambda=y $. If we want to respect our condition $x<y<z$, we find $y=6$. 
In conclusion, $x=4$, $y=6$ and $z=7$ is a solution.
Please note that I have edited this answer because I made a mistake forgetting about the conditions.
