Apart from the slightly unusual nomenclature, this is Shamir's secret-sharing scheme with $n=6$ and $k=3$ (i.e., the secret is shared into six pieces, any three of which can be combined to retrieve the secret).
In this case, the pieces are as follows:
$$\begin{align}(x_0, y_0) &= (10, 25) \\
(x_1, y_1) &= (20, 405) \\
(x_2, y_2) &= (30, 272) \\
(x_3, y_3) &= (40, 129) \\
(x_4, y_4) &= (50, 479) \\
(x_5, y_5) &= (60, 316)\end{align}$$
Only three of these are needed to retrieve the secret. As described in the Wikipedia article, this is done by calculating a set of Lagrange basis polynomials:
$$\begin{align}\ell_0 &= \frac{x - x_1}{x_0 - x_1} \cdot \frac{x - x_2}{x_0 - x_2} = \frac{x - 20}{10 - 20} \cdot \frac{x - 30}{10 - 30} = \frac{x^2}{200} - \frac{x}{4} + 3 \\
\ell_1 &= \frac{x - x_0}{x_1 - x_0} \cdot \frac{x - x_2}{x_1 - x_2} = \frac{x - 10}{20 - 10} \cdot \frac{x - 30}{20 - 30} = \frac{-x^2}{100} + \frac{2x}{5} - 3 \\
\ell_2 &= \frac{x - x_0}{x_2 - x_0} \cdot \frac{x - x_1}{x_2 - x_1} = \frac{x - 10}{30 - 10} \cdot \frac{x - 20}{30 - 20} = \frac{x^2}{200} - \frac{3x}{20} + 1
\end{align}$$
All you need from these polynomials are the constant terms, which are multiplied by the corresponding $y$ terms to give you the secret value:
$$\begin{align}s &= (3 \cdot 25 - 3 \cdot 405 + 1 \cdot 272) \pmod{503} \\
&= 138 \end{align}$$