jmc

Usually when we apply Simon's Favorite Factoring Trick, we have two variables. Maybe we can find an adaptation for three variables. We notice that four of the terms on the left hand side have a factor of $z$ in them, so we can factor it out as: $$z(6xy+21x+2y+7)+30xy+105x+10y=812.$$This looks promising! Add $35$ to each side and continue factoring: $$\begin{array}{rl}z(6xy+21x+2y+7)+30xy+105x+10y+35& =812+35\Rightarrow \\ z(6xy+21x+2y+7)+5(6xy+21x+2y+7)& =812+35\Rightarrow \\ (z+5)(6xy+21x+2y+7)& =847.\end{array}$$Now we can proceed with the two-variable version of Simon's Favorite Factoring Trick on the remaining four-term factor: $$\begin{array}{rl}(z+5)(3x(2y+7)+2y+7)& =847\Rightarrow \\ (z+5)(3x+1)(2y+7)& =847.\end{array}$$The prime factorization of $847$ is $7\cdot 11^2$. We must find $3$ numbers which multiply to $847$ and assign them to $z+5$, $3x+1$, and $2y+7$. We know none of the factors can be negative, since then we would have a negative solution for $x$, $y$ or $z$, which must be positive numbers. Similarly, no factor can be $1$ because that would give either $z=-4$, $x=0$, or $y=-3$, none of which is allowable. There are only $3$ non-one factors which multiply to $847$, so in some order our three factors must be $7$, $11$, and $11$.

We examine the $3x+1$ term. If this factor is equal to $11$, then $x=\frac{10}{3}$, which is not an integer. So $3x+1=7$ and $x=2$. The remaining factors must equal $11$. Setting $2y+7=11$ gives $y=2$, and setting $z+5=11$ gives $z=6$. Thus $x+y+z=2+2+6=\boxed{10}$.