Determine the fraction of land remaining after the son's share.

The son was given $\frac{1}{3}$ of the land.
The remaining land is the total land (represented by 1) minus the son's share.

$$\text{Remaining land after son} = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$$

The daughter was given $\frac{1}{4}$ of the remaining land.

$$\text{Daughter's share} = \frac{1}{4} \times \left(\frac{2}{3}\right) = \frac{2}{12} = \frac{1}{6}$$

The land remaining for the parents is the total land minus the son's share and the daughter's share.

$$\text{Parents' share} = 1 - \text{Son's share} - \text{Daughter's share}$$ $$\text{Parents' share} = 1 - \frac{1}{3} - \frac{1}{6}$$

To subtract these fractions, find a common denominator, which is 6.

$$\text{Parents' share} = \frac{6}{6} - \frac{2}{6} - \frac{1}{6}$$ $$\text{Parents' share} = \frac{6 - 2 - 1}{6} = \frac{3}{6} = \frac{1}{2}$$

Alternatively, after the son took $\frac{1}{3}$, $\frac{2}{3}$ of the land remained. The daughter took $\frac{1}{4}$ of this remaining land. This means $1 - \frac{1}{4} = \frac{3}{4}$ of that remaining land was left for the parents.

$$\text{Parents' share} = \frac{3}{4} \times \left(\frac{2}{3}\right) = \frac{6}{12} = \frac{1}{2}$$

The fraction of the land that remained for the parents is $\frac{1}{2}$.