The [product](https://artofproblemsolving.com/wiki/index.php/Product) $N$ of three [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) is $6$ times their [sum](https://artofproblemsolving.com/wiki/index.php/Sum), and one of the [integers](https://artofproblemsolving.com/wiki/index.php/Integer) is the sum of the other two. Find the sum of all possible values of $N$.
Let the three integers be $a, b, c$.  $N = abc = 6(a + b + c)$ and $c = a + b$.  Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$.  Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ is $12 \cdot (13 + 8 + 7) = 12(28) = \boxed{336}$.