10. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants, and $P(1)=2004, P(2)=$ $4008, P(3)=6012$, then the value of $\frac{1}{4}[P(11)+P(-7)]$ is $\qquad$ .
10. Let $Q(x)=P(x)-2004 x$, then $Q(1)=Q(2)=Q(3)=0$.

Thus, $Q(x)=(x-1)(x-2)(x-3)(x-r)$,
Therefore, by $P(x)=Q(x)+2004 x$, we get
$$
\frac{1}{4}[P(11)+P(-7)]=\frac{1}{4}[Q(11)+2004 \times 11+Q(-7)+2004 \times(-7)]=\frac{1}{4}[Q(11)+Q(-7)]+
$$
2004.

On the other hand: $Q(11)+Q(-7)=10 \cdot 9 \cdot 8 \cdot(11-7)+(-8)(-9)(-10)(-7)-10 \cdot 9 \cdot 8 \cdot 18=$ 12960,
Thus, $\frac{1}{4}[P(11)+P(-7)]=3240+2004=5244$. Fill in 5244.