The quantity in question is seen to be a multiple of 19 as follows:
$$7^{2024} + 7^{2025} + 7^{2026} = 7^{2024} (1 + 7 + 7^2) = 7^{2024} \cdot 57 = 7^{2024} \cdot 3 \cdot 19.$$
Therefore the remainder when it is divided by 19 is 0.

Working modulo 19, $7^0 = 1$, $7^1 = 7 \cdot 1 = 7$, $7^2 = 7 \cdot 7 = 49 \equiv 11$, and $7^3 \equiv 7 \cdot 11 = 77 \equiv 1$.
Therefore the remainders when dividing successive nonegative powers of 7 by 19 repeat with period 3.
The sum of any three consecutive remainders is therefore $1 + 7 + 11 = 19$, so the remainder when the given sum is divided by 19 is 0.