Quadrics intersecting the twisted cubic and a line. I am trying to understand the determinantal approach on Harris book "Algebraic Geometry: A first course" on proving that the intersection of two quadrics containing the twisted cubic in $\mathbb{P}^3$ is the twisted cubic itself and a line (Pg. 110, 111).
The twisted cubic can be described as the zero locus of the $2 \times 2$ minors of the matrix
$$\left(\begin{array}
xx_0 & x_1 & x_2 \\ x_1 & x_2 & x_3
\end{array}\right)$$
If $\lambda = [\lambda_0 : \lambda_1 : \lambda_2] \in \mathbb{P}^2$ any quadric containing the twisted cubic can be neatly described as the zero locus of 
$$\left|\begin{array}
xx_0 & x_1 & x_2 \\ x_1 & x_2 & x_3 \\ \lambda_0 & \lambda_1 & \lambda_2
\end{array}\right|$$
Now,  given $[\mu_0 : \mu_1 : \mu_2] \neq [\lambda_0 : \lambda_1 : \lambda_2]$, Harris then claims that the intersection of two such quadrics away from the twisted cubic is the rank $\le 2$ locus of 
$$\left(\begin{array}
xx_0 & x_1 & x_2 \\ x_1 & x_2 & x_3 \\ \lambda_0 & \lambda_1 & \lambda_2 \\ \mu_0 & \mu_1 & \mu_2
\end{array}\right)$$
i.e. the zero locus of the $3\times 3$ minors of the matrix above, which gives the equations of the quadrics and
$$\left|\begin{array}
xx_0 & x_1 & x_2 \\ \lambda_0 & \lambda_1 & \lambda_2 \\ \mu_0 & \mu_1 & \mu_2
\end{array}\right|= \left|\begin{array}
xx_1 & x_2 & x_3 \\ \lambda_0 & \lambda_1 & \lambda_2 \\ \mu_0 & \mu_1 & \mu_2
\end{array}\right|=0$$
which are equations defining a line.
But why is that? I don't see how we can know a priori that this line resides in the intersection of the quadrics. Or even how we know a priori that the zero locus of these minors will give the part of the intersection which is not in the twisted cubic.
 A: This is my first time posting a maths answer - hope it's okay to answer such an old question! Anyway I am working through Harris's book and was scouring the internet for a solution to this (and related questions Exercise 1.11 Harris Algebraic Geometry: A First Course, An exercise involving the twisted cubic, A problem in Harris: Equations for the twisted cubic) so I feel it's worthwhile having one posted up! This is based on notes by James Mckernan I found through some deep googling which may or may not be meant to be publicly available, I don't know :\
It turns out that each of the conditions or restrictions we impose can be seen as equivalent to specifying the level of in/dependence between various rows that appear in our various matrices. I would note that the line we obtain may intersect the cubic at one or two points -- i.e. Harris's remark that this is “the locus outside $C$” is not meant to be taken too strictly. My source for this solution states that in fact it will intersect the cubic but didn't give details. However if you take $\lambda=[1:0:0]$, $\mu=[0:1:0]$ we obtain the line given by $x_2=x_3=0$ which intersects the cubic at $[1:0:0:0]$.
First showing the line lies in the intersection of the quadrics: As $\lambda$ and $\mu$ are independent, then the determinant form of the equations for our line are the same as saying (for a point $p=[x_0:x_1:x_2:x_3]$ lying on the line) that the triples $(x_0, x_1, x_2)$ and $(x_1, x_2, x_3)$ lie in the span of $\lambda$ and $\mu$. But then either $(x_0, x_1, x_2)$ and $(x_1, x_2, x_3)$ are colinear, in which case $p$ lies on the twisted cubic (and hence on the quadrics defined via $\lambda$ and $\mu$), or they are independent, in which case their span is equal to the span of $\lambda$ and $\mu$ and so $\lambda$, $\mu$ lie in the span of $(x_0, x_1, x_2)$ and $(x_1, x_2, x_3)$, which is equivalent to the determinants
$$ \begin{vmatrix}x_0 & x_1 & x_2\\\ x_1 & x_2 & x_3\\\ \lambda_0 & \lambda_1  &\lambda_2\end{vmatrix}=\begin{vmatrix}x_0&x_1&x_2\\\ x_1&x_2&x_3\\\ \mu_0&\mu_1&\mu_2\end{vmatrix}=0\text{.}$$
Working in the other direction, we seek the intersection of the two quadrics lying outside the twisted cubic as follows:
On the one hand, if our point $p=[x_0:x_1:x_2:x_3]$ lies on both quadrics, that is
$$ \begin{vmatrix}x_0 & x_1 & x_2\\\ x_1 & x_2 & x_3\\\ \lambda_0 & \lambda_1  &\lambda_2\end{vmatrix}=\begin{vmatrix}x_0&x_1&x_2\\\ x_1&x_2&x_3\\\ \mu_0&\mu_1&\mu_2\end{vmatrix}=0\text{,}$$
then each of these matrices have rank $< 3$.
On the other hand, if $p$ does not lie on the twisted cubic then some $2\times 2$ minor of $M=\begin{pmatrix}x_0& x_1& x_2 \\\ x_1& x_2& x_3\end{pmatrix}$ does not vanish: i.e. $M$ has rank $2$, and thus $(x_0, x_1, x_2)$ and $(x_1, x_2, x_3)$ are independent.
Then $\lambda$, $\mu$ lie in the row span of M. As $\lambda$ and $\mu$ are independent, then $(x_0, x_1, x_2)$ and $(x_1, x_2, x_3)$ lie in the span of $\lambda$ and $\mu$, which is equivalent to the determinants
$$ \begin{vmatrix}x_0 & x_1 & x_2\\\ \lambda_0 & \lambda_1  &\lambda_2\\\ \mu_0&\mu_1&\mu_2\end{vmatrix}=\begin{vmatrix}x_1&x_2&x_3\\\ \lambda_0 & \lambda_1  &\lambda_2\\\ \mu_0&\mu_1&\mu_2\end{vmatrix}=0\text{.}$$
