Since \[ \gcd(a, ba + 1) = \gcd(a, 1) = 1 \]\[ \gcd(a, ca^2 + ba + 1) = \gcd(a, 1) = 1 \]\[ \gcd(ba + 1, ca^2 + ba + 1) = \gcd(ba + 1, ca^2) \leq \gcd(ba + 1, c) \times \gcd(ba + 1, a)^2 = 1 \] This indicates that \( a, ba + 1, ca^2 + ba + 1 \) are pairwise coprime. Since \( 2021^{2021} = 43^{2021} \times 47^{2021} \), and \( a < ba + 1 < ca^2 + ba + 1 \), it follows that: \[ a = 1 \]\[ ba + 1 = 43^{2021} \]\[ ca^2 + ba + 1 = 47^{2021} \] Thus, solving for \( a, b, c \): \[ (a, b, c) = \boxed{\left(1,\ 43^{2021} - 1,\ 47^{2021} - 43^{2021}\right)} \]