The quadric contains the whole line I am looking at the following exercise: 
Show that, if a quadric contains three points on a straight line, it contains the whole line. 
Deduce that, if $L_1$, $L_2$ and $L_3$ are nonintersecting straight lines in $\mathbb{R}^3$, there is a quadric containing all three lines. 
$$$$ 
A straight line is of the form $\gamma (t)=a+tb$, right? 
Do we use the following equation that defines the quadric? 
$$v^tAv+b^tv+c=0$$ 
What does it mean that the quadric contains three points on a straight line? 
$$$$ 
EDIT: 
I am looking also at the next exercise: 

I have the following: 
Let $L_1, L_2, L_3$ be three 
nonintersecting
straight lines of the first family. From the previous Exercise we have that there is a quadric that contains all the three lines.
We have that each line of the second family , with at most a  finite number of exceptions, intersects each line of the first family.
Let $\tilde{L}$ such a line of the second family.
So $\tilde{L} $ intersects the lines $L_1, L_2, L_3$.
Since the above quadric contains $L_1, L_2, L_3$ we have that the quadric contains three points on $\tilde{L}$. Therefore the quadric contains the whole $\tilde{L}$. So the quadric contains all the lines of the second family, with at most a finite number of exceptions.
So a doubly ruled surface is a quadric surface, or part of a quadric surface.
Is this correct? 
Which quadric surfaces are doubly ruled? 

For the first part, you are on the good track. Plug in $\gamma(t)$ in the equation of a quadric. As an equation in $t$ it is quadratic and by the assumption it has 3 different zeros. That means it is identically zero, or in other words every point on a line is on a quadric too.
