<s>[INST] Given a tetrahedron, whose sides are $AB=3,AC=4,BC=5,AD=6,BD=7,CD=8$ . Find the volume of the tetrahedral $ABCD$ . 
Given a tetrahedron, whose sides are $AB= 3, AC= 4, BC= 5, AD= 6, BD= 7, CD= 8$ . Find the volume of the tetrahedral $ABCD$ .

Assume that the tetrahedral $ABCD$ has its height $DH$ , whose length I will find by using vectors and the following lemma:

Given three real numbers $f, t, u$ so that $f+ t+ u= 1$ and $H$ on the plane $BCD$ . We'll have
$$\overrightarrow{DH}= f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC}$$

From hypothesis
$$\overrightarrow{DH}\cdot \overrightarrow{AB}= 0\Rightarrow \left ( f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC} \right )\left ( \overrightarrow{DA}- \overrightarrow{DB} \right )= 0$$
$$\Rightarrow f\overrightarrow{DA}\cdot \overrightarrow{DB}+ 49t+ u\overrightarrow{DC}\cdot \overrightarrow{DB}- 36f- t\overrightarrow{DB}\cdot \overrightarrow{DA}- u\overrightarrow{DC}\cdot \overrightarrow{DA}= 0$$
$$49t- 36f+ (f- t)\overrightarrow{DA}\cdot \overrightarrow{DB}+ u\overrightarrow{DC}\cdot \overrightarrow{DB}- u\overrightarrow{DC}\cdot \overrightarrow{DA}= 0$$
On the other hand
$$\cos DCA= \frac{64+ 36- 16}{2\cdot 8\cdot 6}= \frac{7}{8}$$
$$\cos ADB= \frac{36+ 49- 9}{2\cdot 6\cdot 7}= \frac{19}{21}$$
$$\cos DBC= \frac{64+ 49- 25}{2\cdot 8\cdot 7}= \frac{11}{14}$$
Therefore
$$\overrightarrow{DA}\cdot \overrightarrow{DB}= 6\cdot 7\cdot \frac{19}{21}= 38$$
$$\overrightarrow{DC}\cdot \overrightarrow{DB}= 8\cdot 7\cdot \frac{11}{14}= 44$$
$$\overrightarrow{DC}\cdot \overrightarrow{DA}= 8\cdot 6\cdot \frac{7}{8}= 42$$
$$\therefore 49t- 36f+ 38(f- t)+ 44u- 42u= 0\Rightarrow 2f+ 11t+ 2u= 0$$
Similarly
$$\overrightarrow{DH}\cdot \overrightarrow{BC}= 0\Rightarrow \left ( f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC} \right )\left ( \overrightarrow{DC}- \overrightarrow{DB} \right )= 0$$
$$\Rightarrow f\left ( \overrightarrow{DA}\cdot \overrightarrow{DC}- \overrightarrow{DA}\cdot \overrightarrow{DB} \right )+ t\overrightarrow{DB}\cdot \overrightarrow{DC}- t\left | \overrightarrow{DB} \right |^{2}+ u\left | \overrightarrow{DC} \right |^{2}- u\overrightarrow{DC}\cdot \overrightarrow{DB}= 0$$
$$\Rightarrow 4f+ 44t- 49t+ 64u- 44u= 0\Rightarrow 4f- 5t+ 20u= 0$$
A solution to the system of linear equations given by
$$f= \frac{115}{72}, t= -\frac{2}{9}, u= -\frac{3}{8}$$
$$\Rightarrow \overrightarrow{DH}= \frac{115}{72}\overrightarrow{DA}- \frac{2}{9}\overrightarrow{DB}- \frac{3}{8}\overrightarrow{DC}$$
$$\Rightarrow \left | \overrightarrow{DH} \right |^{2}= \frac{115^{2}}{75^{2}}\cdot 36+ \frac{4}{81}\cdot 49+ \frac{9}{64}\cdot 64- \frac{115\cdot 4}{72\cdot 9}\cdot 38+ \frac{2\cdot 6}{9\cdot 8}\cdot 44- \frac{115\cdot 6}{72\cdot 8}\cdot 42= \frac{1199}{36}$$
Is there any way to find the length of $DH$ ? Thanks a real lot.
 [/INST] The volume of the pyramid in terms of its side lengths
can be found as
\begin{align}
V&=
\frac1{12}\,
\left(
4\, u^2\, v^2\, w^2+(u^2+v^2-c^2)\, (v^2+w^2-a^2)\, (u^2+w^2-b^2)
\right.
\\
&\phantom{=}
\left.
-u^2\, (v^2+w^2-a^2)^2-v^2\, (u^2+w^2-b^2)^2-w^2\, (u^2+v^2-c^2)^2
\right)^{1/2}
,
\end{align}
where
\begin{align} 
a&=|BC|
,\quad 
b=|AC|
,\quad 
c=|AB|
,\\ 
u&=|AD|
,\quad 
v=|BD|
,\quad 
w=|CD|
.
\end{align}
For
$|AB|=3, |AC|=4, |BC|=5, |AD|=6, |BD|=7, |CD|=8$,
\begin{align} 
V&=\tfrac13\,\sqrt{1199}
\approx 11.542
.
\end{align}
See also
Cayley-MengerDeterminant,
as advised in
this answer.
Check-in:
one of instances of such tetrahedron can be
presented with the coordinates of vertices as follows:
\begin{align} 
A &= (0, 0, 0)
,\\
B &= (-3, 0, 0)
,\\
C &= (0, 4, 0)
,\\
D &= (\tfrac23, -\tfrac32, \tfrac16\,\sqrt{1199})
.
\end{align}
</s>