Positive integers

Given the equation $6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = 481$, we want to find positive integer solutions for $x$, $y$, and $z$ such that the equation holds.

Notice that all the terms on the left-hand side are divisible by $5$, so let's divide the entire equation by $5$:

$$5(2xyz + 6xy + 3xz + 2yz) + 2x + 17y + z = 96.$$

Now, we can see that the left-hand side is a sum of terms, each of which is divisible by $5$, except for $2x$, $17y$, and $z$.

For the equation to hold, the left-hand side must be a multiple of $5$, and that means $2x + 17y + z$ must be a multiple of $5$.

Looking at the possible values of $2x + 17y + z$, we notice that $2x + 17y + z$ ranges from $2 + 17 + 1 = 20$ (when $x = 1$, $y = 1$, $z = 1$) to $2 \cdot 6 + 17 \cdot 2 + 6 = 56$ (when $x = 6$, $y = 2$, $z = 6$).

Since $20 \leq 2x + 17y + z \leq 56$, and we want it to be a multiple of $5$, the only possibility is $25$ (which is the only multiple of $5$ in that range).

So, we must have $2x + 17y + z = 25$.

To find the positive integer solutions for $x$, $y$, and $z$ that satisfy this equation, we can try different values for $x$, $y$, and $z$ in a systematic way. Starting from $x = 1$, we can substitute values for $y$ and $z$ that satisfy the equation $17y + z = 25 - 2x$.

We find that $x = 1$, $y = 1$, and $z = 7$ satisfy the equation $2x + 17y + z = 25$.

Therefore, $x + y + z = 1 + 1 + 7 = 9$.

So, the value of $x + y + z$ is $\boxed{9}$.