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.

Let the  coordinate of $D$ be ($x,y,z)$ and $\triangle ABC$ be on $xy$- plane, where $A$ is the origin, and $B$ and $C$ are on the coordinate axes.
Distance of $D$ from $B,A,C$ are
$$\begin{align}
(x-3)^2+y^2+z^2 &=49\\
x^2+y^2+z^2 &=36\\
x^2+(y-4)^2+z^2&=64
\end{align}$$ respectively.
Solving these equations we get the value of $z$ which is the height.
$$V=\frac{1}{3}(\text{area of $\triangle ABC$})\times z$$
