How to check if the lines are coplanar? I have two lines:
$\frac{x-1}{3}=\frac{y+2}{-2}=\frac{z}{1} = L_1$ 
$ \frac{x+1}{4} = \frac{y-3}{1}=\frac{z}{\alpha} =L_2$
How can I find the value of $\alpha$ for which these two lines lie on the same plane?
Just notice that this is not a homework question. I took it from an exam held a few years ago, and got no idea about it.
Thanks! 

$$\begin{align*}(L_1)&\;\;z=\frac{x-1}3\;,\;\;\;\;y=\frac{-2x+2}3-2=\frac{-2x-4}3\\
(L_2)&\;\;z=\frac{\alpha x+\alpha}4\;,\;\;y=\frac{x+13}4\end{align*}$$
Suppose there exists a plane $\,ax+by+cz+d=0\;$ which contains both lines, then we have 
$$ax+b\left(\frac{-2x-4}3\right)+c\left(\frac{x-1}3\right)=ax+b\left(\frac{x+13}4\right)+c\left(\frac{\alpha x+\alpha}4\right)\iff$$
$$\left(\frac{11}{12}b+\frac{3\alpha-4}{12}c\right)x+\left(\frac{55}{12}b+\frac{3\alpha+4}{12}c\right)=0$$
and from here we get
$$\begin{align*}11b&=(-3\alpha+4)c\\
55b&=(-3\alpha-4)c\end{align*}$$
Divide second eq. by first one above and get your value of $\,\alpha=2\,$ ...
