Finding the volume of a $\mathbb R^3$ triangle I have a triangle $ABC$ defined  with the points $A=(2.4,-5.4,6)$, $B=(0,1.1,3.2)$, $C=(-7.6,3,0)$
And I'm asked to find the volume of the solid of $\mathbb R^3$ given by the points between the plane that is made by the points of the triangle and the plaze ${z=0}$ (The same points of the triangle with all the $z=0$)
I think I should use integrals but I don't know from which regions to do the triple integration.
 A: First, write down the inequalities that define the triangle. This is the region you'll be integrating across.
Second, compute the function that tells you how far a point, $(x,y,z)$ is from the plane $z=0$. This will be the function you integrate.
Third, put these two together and try computing the integral.
A: The area of a triangle with vertices $A,B,C$ in $\mathbb{R}^3$ can be obtained using the cross product: $$\mathrm{Surface}=\frac12 \|\overrightarrow{AB}\times \overrightarrow{AC}\|.$$ 
The following doesn't give the volume of the region in the question as noticed by @achillehui
Also, the volume of a tetrahedron defined by four points $A,B,C,D$, can be computed using $$\mathrm{Volume}=\frac16 |\det(\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD})|.$$ 
Edit
Assume that $A',B',C'=C$ are the vertices of the triangle on the plane $z=0.$ Then, we can decompose the figure in two tetrahedrons: one with vertices $A',A,B',C$ and the other with vertices $A,B,B',C.$ Using the formula above one gets the volume.
A: The volume is the absolute value of $\frac{6+3.2+0}3·
\frac{\begin{vmatrix}
2.4&-5.4&1\\
0&1.1&1\\
-7.6&3&1
\end{vmatrix}}2$ .    
The first fraction is the effective altitude of the prism which will be used as one of the factors in the volume calculation. The second fraction is the area of the base.
