[Algebra]
Problem: Let $T$ be a rational number. Let $a, b$, and $c$ be the three solutions of the equation $x^{3}-20 x^{2}+19 x+T=0$. Compute $a^{2}+b^{2}+c^{2}$.

Solution: According to Vieta's formulas, $a+b+c=-(-20)=20$ and $a b+b c+c a=19$. Noting that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)$, it follows that $a^{2}+b^{2}+c^{2}=20^{2}-2 \cdot 19=\mathbf{3 6 2}$. The value of $T$ is irrelevant.

Final answer: 362