= Solution
The <Lagrange interpolation polynomial> basis for nodes $0,2,3$ gives
$$
\boxed{a(x)=\frac{(x-2)(x-3)}6,\qquad
b(x)=-\frac{x(x-3)}2,\qquad
c(x)=\frac{x(x-2)}3.}
$$
Each polynomial is one at its own node and zero at the other two. For a polynomial $f$ of degree at most two, the difference between $f$ and this interpolation polynomial has degree at most two and three distinct roots, so it is identically zero. This also proves uniqueness of the quadratic coefficients.
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