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.
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}