3. Given that $f(x)$ represents a fourth-degree polynomial in $x$, and $f(a)$ represents the value of $f(x)$ when $x=a$. If
$$
\begin{array}{l}
f(1)=f(2)=f(3)=0, \\
f(4)=6, f(5)=72,
\end{array}
$$

then the value of $f(6)$ is ( ).
(A) 200
(B) 300
(C) 400
(D) 600
3. B.

Since $f(1)=f(2)=f(3)=0$, the quartic polynomial $f(x)$ can be set as
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {. }
$$

Given $f(4)=6, f(5)=72$, we have
$$
\left\{\begin{array}{l}
6(4 a+b)=6, \\
24(5 a+b)=72 .
\end{array}\right.
$$

Solving these equations, we get $a=2, b=-7$.
Thus, $f(x)=(x-1)(x-2)(x-3)(2 x-7)$.
Therefore, $f(6)=5 \times 4 \times 3 \times 5=300$.