Let $x$ be the number of $Al(OH)_3$; $y$ the number of $H_2SO_4$; $z$ the number of $AL_2(SO_4)_3$, and $w$ the number of $H_2O$. Looking at the number of $Al$, you get $x = 2z$. Looking at $O$, you get $3x + 4y = 12z + w$. Looking at $H$ you get $3x + 2y = 2w$; and looking at $S$ you get $y = 3z$. That looks like what you are getting from Wolfram, except you have the wrong signs for $z$ and $w$; unless you are interpreting the first two entries to represent the "unknowns", and the last two to represent the "solutions". I would translate into equations the usual way.
What you have is the following system of linear equations:
x & & & & -2z & & & = & 0\\
3x & + & 4y & - & 12z & - & w & = & 0\\
3x & + & 2y & & & - & 2w & = & 0\\
& & y & - & 3z & & & = & 0
This leads (after either some back-substitution from the first and last equations into the second and third, or some easy row reduction) to $x=2z$, $y=3z$, and $6z=w$. Since you only want positive integer solutions, setting $z=1$ gives $x=2$, $y=3$, and $w=6$, yielding the smallest solution:
$$2 Al(OH)_3 + 3H_2SO_4 \longrightarrow Al_2(SO_4)_3 + 6H_2O.$$