Problem to Brilliant! 7

Using Newton's Sums method, we know that:

$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$

$\Rightarrow ab+bc+ca = \dfrac {(a+b+c)^2 - a^2+b^2+c^2}{2} = \dfrac {81 - 99}{2} = -9$

Also that:

$a^3+b^3+c^3 = (a+b+c)(a^2+b^2+c^2- ab-bc-ca)+3abc$

$\Rightarrow abc = \dfrac {a^3+b^3+c^3 - (a+b+c)(a^2+b^2+c^2- ab-bc-ca)}{3}$

$\quad \quad \quad = \dfrac {999-9(99+9)}{3} = \dfrac {999-9(99+9)}{3} = \dfrac {27}{3} = 9$

And:

$a^4+b^4+c^4 = (a+b+c)(a^3+b^3+c^3)$

$\quad \quad \quad \quad \quad \quad \space - (ab+bc+ca)(a^2+b^2+c^2) +abc(a+b+c)$

$\quad \quad \quad \quad \quad \quad = 9(999)+9(99) +9(9) = 9963$

Therefore,

$a^5+b^5+c^5 = (a+b+c)(a^4+b^4+c^4)$

$\quad \quad \quad \quad \quad \quad \quad + (ab+bc+ca)(a^3+b^3+c^3) +abc(a^2+b^2+c^2)$

$\quad \quad \quad \quad \quad \space = 9 (9963) +9(999) +9(99) = 99549 = \alpha$

And $\alpha \mod {78} = 99549 \mod {78} = \boxed{21}$