There is a unique quadric through three disjoint lines There is a classical exercise that three disjoint lines in $\mathbb{P}^3$ are contained in a quadric surface $Q$. The existence is trivial. Every quadric in $\mathbb{P}^3$ is determined by nine coefficients and if three distinct points of a line lie on $Q$ then the whole line lie on $Q$. Thus we obtain the system of 9 linear conditions on $Q$. But why such a quadric is unique? To be more precise, why these linear equations are linearly independent?
 A: There exists exactly one quadric containing  three mutually disjoint lines in $\mathbb P^3$:
Indeed you may assume that these lines are 
$$L_1: x_0=x_1=0\quad \quad L_2:x_2=x_3=0\quad\quad  L_3: x_0=x_2 ,\:x_1=x_3$$ 
and then the unique quadric containing these three line  $L_i$  has the equation 
$$x_0x_3-x_1x_2=0$$ 
This is easy to  check  by writing that the polynomial 
$$ax_0^2+bx_0x_1+\dots+jx_3^2$$
vanishes on the $L_i$'s.
A: Mutually skew lines $a,b$ and $c$ lie in the same ruling, and the quadric is the union of the lines of the other ruling. There exists exactly one line containing a given point $p \in a$ and intersecting $b$ and $c$, so the quadric is unique as the union of all the lines intersecting with $a,b$ and $c$.
A: You already have two good answers; let me give a bad one. Here "bad" means "uses more macinery than necessary"; I am writing it anyway in case you like the approach.
Let $Q$ be the quadric containing our three lines $L_1, L_2, L_3$. Now for any other quadric $Q' \subset \mathbf P^3$, the intersection $Q \cap Q'$ is a quartic curve. 
Suppose now there exists another $Q'$ which also contains $L_1$, $L_2$, $L_3$. Then $Q \cap Q' = L_1 \cup L_2 \cup L_3 \cup L$ for some other line $L$. 
Now we use some machinery: intersection numbers. (Assuming we're working over $\mathbf C$, you can argue in terms of singular cohomology.) Fix one of our lines, say $L_1$. In $\mathbf P^3$ we know $L_1 \cdot Q' =2$, so on $Q$ we know that $L_1 \cdot (L_1 \cup L_2 \cup L_2 \cup L)=2$. But $L_1 \cdot L_j = 0$ for $j=1,2,3$ since these three lines are all in one ruling, and $L_1 \cdot L \leq 1$. This is a contradiction.
