$\text{ Suppose }a,b\text{ and }c\text{ are positive integers with }a<b<c\text{ such that }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.\text{ The value of }(a+b+c)\text{ is }$

$\begin{array}{l}\text{ We must have }\frac{1}{a}<1,\text{ so }a>1\\ \text{ Since }\frac{1}{a}>\frac{1}{b}>\frac{1}{c}\text{ . }\\ \Rightarrow \frac{1}{a}>\frac{1}{3}\\ \Rightarrow a<3\Rightarrow a=2\\ \Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{2}\text{ where }2<b<c\\ \text{ Similarly }\frac{1}{b}>\frac{1}{4}\\ \text{ so }b<4\Rightarrow b=3\end{array}$

$\begin{array}{l}\text{ Now }c=6\text{ satisfies the equation }\\ \Rightarrow a+b+c=11\end{array}$