Consecutive is unpredictable (Part 2)

Let $$\large{\color{red}a\color{blue}b\color{purple}c}$$ be the last three digits of $$\large{1^{2}+2^{3}+3^{4}+4^{5}+5^{6}+...+998^{999}+999^{1000}}$$.

Find $$\large{\color{red}a\color{blue}b\color{purple}c}$$ $$mod$$ $$69$$.

Details:

The exponents are in a consecutive order of the positive integers: $$2, 3, 4, ..., 998, 999, 1000$$

$$\large{\color{red}a\color{blue}b\color{purple}c}$$ is three different digits, like $$346$$, not algebraic expression like $$\large{\color{red}a \times\color{blue}b\times\color{purple}c}$$ $$(3\times4\times6)$$

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