Intersection of a smooth projective variety and a plane Let $X \subset P^n$ be an irreducible smooth complex projective
variety embedded in the $n$-dimensional projective space.
Let $k$ be the dimension of $X$ and $d$ its degree.
Let $L \subset P^n$ be a linear subspace of dimension $n-k$ 
and $Z=L \cap X$. Assume that
(a) $X$ is not contained in any hyperplane of $P^n$ and
(b) $Z$ is finite of cardinality $d$.
Question: Is it true that $Z$ spans $L$?
Comment: I was told that this is true if $X$ is ACM 
(arithmetically Cohen-Macaulay). A reference for this 
would be appreciated.
 A: It is true that $Z$ spans $L$ — even if $X$ isn't ACM.   You can also allow $X$ to be singular (but you do need $X$ irreducible and non-degenerate, of course).  To illustrate one of the main ideas it is useful to first look at the case when $X$ 
is a curve.
If $X$ is a curve.  Let $M$ be the span of $Z$ and suppose that $M\neq L$.  (In the curve case, $L$ will be a hyperplane).  Let $p$ be any point of $X$ outside of $Z$ and let $H$ be any hyperplane containing $M$ and $p$.  Then $H\cap X$ contains at least $d+1$ points, so by Bezout's theorem the intersection cannot be zero dimensional.  Since $X$ is irreducible and one dimensional, this means that the intersection must be all of $X$, so $X$ is contained in $H$, contrary to hypothesis.     
The general case.
The idea when $k\geqslant 2$ is to show that if $H$ is a general hyperplane containing $L$ then $H \cap X$ is irreducible and non-degenerate (i.e, the intersection $H\cap X$ is not contained in a smaller linear space of $H$).  But now all dimensions have been reduced by $1$, and so iterating this procedure reduces us to the curve case, which we've already solved.
To set this up, note that hyperplanes in $\mathbb{P}^n$ containing $L$ are parameterized by a $\mathbb{P}^{k-1}$ (If $V$ is the underlying vector space of $\mathbb{P}^{n}$, $W$ the underlying vector space of $L$, then the hyperplanes are parameterized by the projectivization of $(V/W)^{*}$).  We'll use $H$ to refer both to a point of $\mathbb{P}^{k-1}$ and the corresponding hyperplane in $\mathbb{P}^n$ containing $L$. Define $\Gamma\subset \mathbb{P}^{k-1}\times (X\setminus Z)$ to be the set
$$\Gamma = \left\{(H,p) \mid p\in H\right\}$$
i.e, the pairs $(H,p)$ so that $H$ is a hyperplane containing $L$, and $p$ a point of $H\cap X$ not on $Z$.  
If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$.  In other words,  $\Gamma$ is a $\mathbb{P}^{k-2}$ bundle over $X\setminus Z$. (This fibration is where we use $k\geqslant 2$.) Since $X\setminus Z$ is irreducible this implies that $\Gamma$ is irreducible. 
Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$.  Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is.  For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$.
Now let $q$ be any point of $Z$.  Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$.  Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$.  Finally, since $Z$ is the intersection of all $X\cap H$ for $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$.   Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres.  Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible. 
(The intuitive reason for this implication is that, generically over $\mathbb{P}^{k-1}$ the section lets us pick out precisely one irreducible component of the fibre.  The union of these components gives us a subset of $\overline{\Gamma}$ which has the same dimension as $\overline{\Gamma}$, and hence whose closure must be all of $\overline{\Gamma}$ by irreducibility.  But if there is more than one component in a general fibre, this is a contradiction, thus the general fibre must be irreducible.  To make this intuitive construction rigorous requires passing to the normalization of $\overline{\Gamma}$ and then looking at the Stein factorization of the map from the normalization to $\mathbb{P}^{k-1}$.  The section gives a generic section of the finite part of the Stein factorization,  and that allows one to construct the ``union of the components containing the section''.)
Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate.  Let  $Y=H\cap X$, so that $Y$ is a variety of degree $d$ and dimension $k-1$.  Let $M$ be the span of $Y$.  If $M\neq H$ then pick any point  $p\in X\setminus Y$ and let $H'$ be any hyperplane containing $M$ and $p$.  Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Y'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Y'$ have dimension $k-1$) and degree $d$.  But $Y$ is therefore a component of $Y'$, and the equality of degrees tells us that $Y'$ can't have any other components so we must have $Y'=Y$.  This contradicts the fact that $p\in Y'$ and $p\not\in Y$.
Together this shows the required inductive step:  If $H$ is a general hyperplane containing $L$ then $H\cap X$ is irreducible and non-degenerate. 
Other remarks. I'm guessing from the setup of the question that you want to apply the  result for a particular $L$ that you have chosen.   If, in the application, you're allowed to pick a general $L$ then you can say something stronger.  The classical uniform position principle (where ''classical'' in this case means ''established by Joe Harris in the 80's'') states that for a general subspace $L$ of dimension $n-k$ the finite set of $d$-points in $Z=L\cap X$ have the property that any subset of $r+1$ of the points (with $r\leqslant n-k$) span a $\mathbb{P}^{r}$.  Picking $r=n-k$, this means that any subset of $n-k+1$ points of $Z$ spans all of $L$, and so in particular $Z$ spans $L$.  (Note that $d\geqslant n-k+1$; for instance, as a consequence of the argument above: if $d < n-k+1$ then the $d$ points of $Z$ would never span $L$.)
