Example 10 Let $p(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants, and $p(1)=1993$, $p(2)=3986$, $p(3)=5979$. Try to calculate $\frac{1}{4}[p(11)+p(-7)]$. \quad (1993 Macau Olympiad Question)
Let $Q(x) = p(x) - 1993x$, then $Q(1) = Q(2) = Q(3) = 0$.
Thus, $Q(x) = (x-1)(x-2)(x-3)(x-r)$.
Therefore, from $p(x) = Q(x) + 1993x$, we have
$$
\begin{aligned}
\frac{1}{4}[p(11) + p(-7)] & = \frac{1}{4}[Q(11) + 1993 \cdot 11 + Q(-7) + 1993 \cdot (-7)] \\
& = \frac{1}{4}[Q(11) + Q(-7)] + 1993.
\end{aligned}
$$

But $Q(11) = 10 \cdot 9 \cdot 8 \cdot (11 - r)$,
$$
Q(-7) = (-8) \cdot (-9) \cdot (-10) \cdot (-7 - r),
$$

So $Q(11) + Q(-7) = 10 \cdot 9 \cdot 8 \cdot 18 = 12960$.
Hence $\frac{1}{4}[p(11) + p(-7)] = 3240 + 1993 = 5233$.