3. Given three vertices of a cube are
$$
P(4,7,4), Q(7,11,4), R(11,8,9) \text {. }
$$

then the coordinates of the center of the cube are $\qquad$
3. $\left(\frac{15}{2}, \frac{15}{2}, \frac{13}{2}\right)$.

Notice that,
$$
\begin{array}{l}
P Q=\sqrt{(4-7)^{2}+(7-11)^{2}+(4-4)^{2}}=5, \\
Q R=\sqrt{(7-11)^{2}+(11-8)^{2}+(4-9)^{2}}=5 \sqrt{2}, \\
R P=\sqrt{(11-4)^{2}+(8-7)^{2}+(9-4)^{2}}=5 \sqrt{3} .
\end{array}
$$

Thus, $P Q: Q R: R P=1: \sqrt{2}: \sqrt{3}$. 
Therefore, $R$ and $P$ are the vertices of the body diagonal of a cube. Hence, the center of the cube is the midpoint of $R P$, which is $\left(\frac{15}{2}, \frac{15}{2}, \frac{13}{2}\right)$.