Finding the volume of a pyramid (the vector way) The problem
I have 4 points
$
P \; (-1,2,0) \\
Q \; (2,1,3) \\
R \; (1,0,1) \\
S \; (3,-2,3)
$
and I want to find the volume of a pyramid. What I'm most concerned here is the appropriate strategy to go about this.
What I've tried


*

*Since the volume is calculated by $\mbox{base} \cdot \mbox{height} \times \frac{1}{3}$ I first find the base

*The base in this case $\frac{1}{2}\lVert PQ\times QR \rVert$

*Next the height vector will co collinear with the normal vector, so I find that from the cross product ${\bf n} = PQ\times QR$


By this point, I've decided that $S$ is the top of the pyramid, and $PQR$ is the base. What I struggle with is finding the distance between the top and base, I.e. the height.
In my mind, that has to be a scaled version of the $\bf n$, but how do I find the scale? Or is my approach broken? 
 A: Hint: The volume of a Parallelepiped is given by the determinant of three of its spanning vectors. (That is $Q-P$, $R-P$ and $S-P$, for instance)
A: $\newcommand{\+}{^{\dagger}}%
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'Move' $S$ to the origin of coordinates. We get the points $\vec{p} = \pars{-2,4,-3}$, $\vec{q} = \pars{-1,3,0}$, $\vec{r} = \pars{-2,2,-2}$ and $\vec{s} = \pars{0,0,0}$ which correspond to the "old" points $P$, $Q$, $R$ y $S$, respectively. With the 'new' points:
$$
V = \int_{V}\dd V = \int_{V}{\nabla\cdot\vec{r} \over 3}\,\dd V
= {1 \over 3}\int_{S}\vec{r}\cdot\dd\vec{S}\tag{1}
$$
where we used Gauss's theorem. The integration over the 'walls' vanishes out since $\vec{r}\cdot\dd\vec{S} = 0$. There remains an integration over a triangle
$\pars{~\mbox{the pyramid base}\ B~}$which vertexes at $\vec{p}$, $\vec{q}$ and $\vec{r}$.
$\pars{1}$ is reduced to
$\ds{V = {1 \over 3}\int_{B}\vec{r}\cdot\hat{n}\,\dd S}$ where $\hat{n}$ is a perpendicular unit vector to $B$. $\pars{~\dd S \equiv \verts{\dd\vec{S}}~}$:
$$
\hat{n}
\equiv
{\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}}
 \over
 \verts{\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}}}}\,,\quad
\left\vert%
\begin{array}{rcl}
\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}}
& = &
\pars{-1,1,-3}\times\pars{0,2,-1}
\\
& = & \pars{5,-1,-2}
\\[2mm]
\verts{\pars{\vec{p} - \vec{q}} \times \pars{\vec{p} - \vec{r}}}
& = &
\root{30}
\\[2mm]
\hat{n} & = & {1 \over \root{30}}\,\pars{5,-1,-2}
\end{array}\right.
$$
Then,
$$
V = {\root{30} \over 90}\int_{B}\pars{5x - y -2z}\dd S
$$
Can you take from here ?
