5. In the tetrahedron $ABCD$,
\[
\begin{array}{l}
AB=1, BC=2 \sqrt{6}, CD=5, \\
DA=7, AC=5, BD=7 .
\end{array}
\]

Then its volume is $\qquad$
5. $\frac{\sqrt{66}}{2}$.

From $A B^{2}+B C^{2}=1^{2}+(2 \sqrt{6})^{2}=25=A C^{2}$, $B C^{2}+C D^{2}=(2 \sqrt{6})^{2}+5^{2}=49=B D^{2}$, we know that $A B \perp B C, B C \perp C D$.
As shown in Figure 2, construct rectangle $A B C E$.
Thus $V$ : volume of $D-A B C$
Since $A E \perp E C$,
$A E \perp C D$,

we know that $A E \perp$ plane $E C D$.
Also, from $A E \perp E D$ and

the Pythagorean theorem, we get $D E=5$.
Furthermore, since the sides of $\triangle E C D$ are $1, 5, 5$, its area is $\frac{3 \sqrt{11}}{4}$.
Thus, $V_{\text {tetrahedron } A-E C D}=\frac{1}{3} \times 2 \sqrt{6} \times \frac{3 \sqrt{11}}{4}=\frac{\sqrt{66}}{2}$ $\Rightarrow V_{\text {prism } A B C D}=\frac{\sqrt{66}}{2}$.