Problem 11.3. On the coordinate plane, all points $(x, y)$ such that $x$ and $y$ are integers satisfying the inequalities $0 \leqslant x \leqslant 2$ and $0 \leqslant y \leqslant 26$ are marked. How many lines exist that pass through exactly 3 of the marked points?
Answer: 365.

Solution. All marked points are located on three vertical lines $x=0, x=1, x=2$. Let's call these lines the left, middle, and right lines, respectively.

Consider any three marked points lying on one line. If at least two of them lie on a vertical line, then there are more than 3 marked points on such a line, which does not fit our criteria. Therefore, on each of the three vertical lines, exactly one point must be chosen.

Notice that when choosing point $A(0, a)$ on the left line and point $B(2, b)$ on the right line, the point on the middle line is uniquely determined - it is the midpoint of segment $AB$, and it has coordinates $(1, \frac{a+b}{2})$. It has integer coordinates if and only if the numbers $a$ and $b$ have the same parity. Consequently, the number of lines passing through exactly three marked points is equal to the number of pairs $(a, b)$ for which $a$ and $b$ are integers of the same parity from 0 to 26 inclusive (it is clear that each triplet of integers $\left(a, \frac{a+b}{2}, b\right)$ corresponds to its own line containing exactly three marked points). On the segment $[0 ; 26]$, there are 14 even and 13 odd numbers, so the answer to the problem is the number $14 \cdot 14 + 13 \cdot 13 = 365$.