Trigonometry - Calculating the pyramid volume The problem:

There be the points $P_0(0,0,0)$, $P_1(1,1,1)$, $P_2(2,-1,2)$ and $P_3(3,0,1)$. Calculate the volume of the pyramid.

Now I assumed the base of the pyramid is a triangle, with points $P_1$, $P_2$ and $P_3$.
So I know $\underline u:=\overrightarrow{P_1P_2} = <1, -2, 1>$ and $\underline v:=\overrightarrow{P_1P_3}=<2,-1,0>$.
I calculated the angle between $\underline u$ and $\underline v$:
$$\cos\theta:=\frac{\underline{u}\cdot\underline{v}}{||\underline{u}||\,||\underline{v}||}=\frac{4}{\sqrt{30}}\Longrightarrow \theta=43.0887^\circ$$
$$S_\Delta=\frac{||\underline{u}||\,||\underline{v}||\sin\theta}{2}=1.87$$
and I thought the pyramid's height to be $\,2\,$ , so the volume is $\,\,\displaystyle{V=\frac{S_\Delta\cdot 2}{3}=1.246}$
I believe I'm somewhat wrong with this.
Help will be much appreciated!

This problem is relatively simple because one of your points is at the origin.  The volume of your pyramid is then equal to $1/6$ of the determinant of the matrix formed by the other points:
$$V = \frac{1}{6} \left |\begin{array}\\1&1&1\\2&-1&2\\3&0&1 \end{array} \right |$$
Evaluating the determinant, I get $V=1$.
