Angular velocity by velocities of 3 particles of the solid Velocities of 3 particles of the solid, which don't lie on a single straight line, $V_1, V_2, V_3$ are given (as vector-functions). Radius-vectors $r_1, r_2$ from third particle to first and second  are given aswell. How could I find the angular velocity $w$ of the solid?
I  tried to solve this problem using Euler's theorem : $V_2=V_3+[w \times r_2]$, $V_1=V_3+[w \times r_1]$. 
After this step I tried to consider different cases: if $V_1 $ is not collinear to $V_2$ we could write $w = k*[(V_2-V_3) \times (V_1-V_3)]$. However, it doesn't really help. The second case is even more difficult to analyze.
Second attempt consisted in solving this system by multiplication (scalar product or vector work) equations by appropriate vectors. However,  I didn't really succeed.
 A: The algebra is not especially nice, but it is just algebra. This is rigid body rotation, taking point 3 as the origin of coordinates,
so effectively 
$$\mathbf{r}_1=\mathbf{R}_1-\mathbf{R}_3, \qquad
\mathbf{r}_2=\mathbf{R}_2-\mathbf{R}_3.
$$
We start as you suggested, and abbreviate
$$
\mathbf{v}_1=\mathbf{V}_1-\mathbf{V}_3, \qquad
\mathbf{v}_2=\mathbf{V}_2-\mathbf{V}_3,
$$
so that
$$
\mathbf{v}_1 = \boldsymbol{\omega}\times\mathbf{r}_1, \qquad
\mathbf{v}_2 = \boldsymbol{\omega}\times\mathbf{r}_2.
$$
Now since the three points are not collinear, we can let
$$
\boldsymbol{\omega} = a\,\mathbf{r}_1 + b\,\mathbf{r}_2
+ c\, \mathbf{r}_1\times\mathbf{r}_2
$$
but we must remember that $\mathbf{r}_1$ and $\mathbf{r}_2$ will not in general be orthogonal.
We can obtain $c$ directly,
from either of the two equivalent equations
\begin{align*}
\mathbf{r}_2\cdot\mathbf{v}_1 
&= \mathbf{r}_2\cdot\boldsymbol{\omega}\times\mathbf{r}_1
= \boldsymbol{\omega}\cdot\mathbf{r}_1\times\mathbf{r}_2
= c |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\mathbf{r}_1\cdot\mathbf{v}_2 
&= \mathbf{r}_1\cdot\boldsymbol{\omega}\times\mathbf{r}_2
= -\boldsymbol{\omega}\cdot\mathbf{r}_1\times\mathbf{r}_2
= -c |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\Rightarrow\quad c&=
\frac{\mathbf{r}_2\cdot\mathbf{v}_1}{|\mathbf{r}_1\times\mathbf{r}_2|^2} 
=
-\frac{\mathbf{r}_1\cdot\mathbf{v}_2}{|\mathbf{r}_1\times\mathbf{r}_2|^2}
\end{align*}
where we took advantage of the properties of the
scalar triple product.
The other coefficients come from scalar products with $\mathbf{r}_1\times\mathbf{r}_2$.
We use the general identity
$$
(\mathbf{A}\times\mathbf{B})\cdot(\mathbf{C}\times\mathbf{D})
=
(\mathbf{A}\cdot\mathbf{C})\,(\mathbf{B}\cdot\mathbf{D}) -
(\mathbf{B}\cdot\mathbf{C})\,(\mathbf{A}\cdot\mathbf{D})
$$
and a special case of this, which we use, is
$|\mathbf{r}_1\times\mathbf{r}_2|^2=|\mathbf{r}_1|^2|\mathbf{r}_2|^2-(\mathbf{r}_1\cdot\mathbf{r}_2)^2$.
\begin{align*}
\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_2 
&= 
(\mathbf{r}_1\times\mathbf{r}_2 ) \cdot (\boldsymbol{\omega}\times\mathbf{r}_2)
\\
&=
\left( a|\mathbf{r}_1|^2 + b(\mathbf{r}_1\cdot\mathbf{r}_2) \right)\,
|\mathbf{r}_2|^2- 
\left( a(\mathbf{r}_1\cdot\mathbf{r}_2) + b|\mathbf{r}_2|^2 \right)\,
(\mathbf{r}_1\cdot\mathbf{r}_2) 
\\
&= a |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\Rightarrow\quad
a&=\frac{\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_2 }{|\mathbf{r}_1\times\mathbf{r}_2|^2}
\\
\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_1 
&= 
(\mathbf{r}_1\times\mathbf{r}_2 ) \cdot (\boldsymbol{\omega}\times\mathbf{r}_1)
\\
&=
\left( a|\mathbf{r}_1|^2 + b(\mathbf{r}_1\cdot\mathbf{r}_2) \right)\,(\mathbf{r}_1\cdot\mathbf{r}_2) - 
\left( a(\mathbf{r}_1\cdot\mathbf{r}_2) + b|\mathbf{r}_2|^2 \right) \,
|\mathbf{r}_1|^2
\\
&= -b |\mathbf{r}_1\times\mathbf{r}_2|^2
\\
\Rightarrow\quad
b &=-\frac{\mathbf{r}_1\times\mathbf{r}_2 \cdot \mathbf{v}_1}{|\mathbf{r}_1\times\mathbf{r}_2|^2}
\end{align*}
I hope I haven't made any slips, you should definitely check!
