Balance the chemical equation: wH2SO4 + xAl(OH)3 -> yH2O + zAl2(SO4)3

On the reactant side, there are $x$ Al atoms in $x\text{Al}(\text{OH})_3$.
On the product side, there are $2z$ Al atoms in $z\text{Al}_2(\text{SO}_4)_3$.
So, $x = 2z$.

On the reactant side, there are $w$ $\text{SO}_4$ groups in $w\text{H}_2\text{SO}_4$.
On the product side, there are $3z$ $\text{SO}_4$ groups in $z\text{Al}_2(\text{SO}_4)_3$.
So, $w = 3z$.

Let's assume $z = 1$ (the simplest whole number).
Then, from Step 1: $x = 2 \times 1 = 2$.
And from Step 2: $w = 3 \times 1 = 3$.

The equation becomes:
$3\text{H}_2\text{SO}_4 + 2\text{Al}(\text{OH})_3 \rightarrow y\text{H}_2\text{O} + 1\text{Al}_2(\text{SO}_4)_3$

Count H atoms on the reactant side:
From $3\text{H}_2\text{SO}_4$: $3 \times 2 = 6$ H atoms.
From $2\text{Al}(\text{OH})_3$: $2 \times 3 = 6$ H atoms.
Total H atoms on reactant side = $6 + 6 = 12$.

Count H atoms on the product side:
From $y\text{H}_2\text{O}$: $y \times 2 = 2y$ H atoms.
Equating H atoms: $12 = 2y \Rightarrow y = 6$.

Count O atoms on the reactant side:
From $3\text{H}_2\text{SO}_4$: $3 \times 4 = 12$ O atoms.
From $2\text{Al}(\text{OH})_3$: $2 \times 3 = 6$ O atoms.
Total O atoms on reactant side = $12 + 6 = 18$.

Count O atoms on the product side:
From $y\text{H}_2\text{O}$ (with $y=6$): $6 \times 1 = 6$ O atoms.
From $1\text{Al}_2(\text{SO}_4)_3$: $1 \times (4 \times 3) = 12$ O atoms.
Total O atoms on product side = $6 + 12 = 18$.
The oxygen atoms are balanced.

The respective values for $w, x, y, z$ are $3, 2, 6, 1$.

A: 2,2,6 and 2
B: 2,2,5 and 1
C: 3,2,5 and 2
D: 3,2,6 and 1

The calculated values match option D.

The final answer is $\text{D}$.