$5 \cdot 24$ The parabola $y=a x^{2}+b x+c$ has its vertex at $(4,2)$. If $(2,0)$ is on the parabola, then $a b c$ equals
(A) -12 .
(B) -6 .
(C) 0 .
(D) 6 .
(E) 12 .
(37th American High School Mathematics Examination, 1986)
[Solution] According to the problem, the parabola passes through the points $(2,0),(4,2),(6$,
$0)$, thus we have
$$
\left\{\begin{array}{l}
16 a+4 b+c=2, \\
4 a+2 b+c=0, \\
36 a+6 b+c=0 .
\end{array}\right.
$$

Solving these equations, we get $a=-\frac{1}{2}, b=4, c=-6$. Therefore, $a b c=12$. Hence, the answer is $(E)$.