Finding a non-degenerated quadric containing three projective lines in $\mathbb{P}^3(\mathbb{C})$

Brute force works without a ton of fuss, since it turns this in to a linear algebra problem: write $$\varphi = ax_0^2+bx_0x_1+cx_0x_2+dx_0x_3+ex_1^2+fx_1x_2+gx_1x_3+hx_2^2+ix_2x_3+jx_3^2$$ and plug in.

• The first line says that $ax_0^2+dx_0x_3+jx_3^2$ must vanish, so $a=d=j=0$.
• The second line says that $(a+b+e)x_0^2+(c+d+f+g)x_0x_2+(h+i+j)x_2^2$ must vanish, so $(a+b+e)=(c+d+f+g)=(h+i+j)=0$.
• The third line says that $ex_1^2+(2f+g)x_1x_3+(4h+2i+j)x_3^2$ must vanish, so $e=(2f+g)=(4h+2i+j)=0$.

After a bit of linear algebra, you'll find that $a=b=d=e=h=i=j=0$ and $f=c$ and $g=-2c$. So your quadric is a multiple of $x_0x_2+x_1x_2-2x_1x_3$, which does indeed vanish on all three lines and is nondegenerate because it's nonsingular (check the Jacobian).

It is true that there's a unique quadric through any triple of disjoint lines in $\Bbb P^3$, but I don't know how to prove that result without doing essentially this again. Maybe someone else will have something smarter to say.