Show that $M\cap N$ is the set of all scalar multiples of vector $a$ in $M \cap N$. 
Show that the set of triples $x \in \mathbb{R}^3$ such that $x_1 - x_2 + 2x_3 = 0$ is a subspace $M$. If $N$ is the similar subspace $\{x : x_1 + x_2 + x_3 = 0\}$, find a nonzero vector $a$ in $M\cap N$. Show that $M\cap N$ is the set $\{xa : x \in \mathbb{R}\}$ of all scalar multiples of $a$. 

Suppose $y \in \mathbb{R}^3$. Then 
\begin{align}
(x_1 - x_2 + 2x_3) + (y_1 - y_2 + 2y_3) &= 0 \\
x_1 + y_1 - x_2 - y_2 + 2(x_3 + y_3) &= 0
\end{align}
For a scalar value $c$, 
\begin{align}
c(x_1 - x_2 + 2x_3) &= 0c \\
cx_1 - cx_2 + 2cx_3 &= 0
\end{align}
Thus the set is a subspace of $M$.
If $N$ is a similar subspace $\{ x : x_1 + x_2 + x_3 = 0\}$ then the intersection $M\cap N$ is the set of triples $x \in \mathbb{R}^3$ such that $x_1 + x_2 + x_3 = 0$. Here I am unsure what "find a nonzero vector $a$ in $M\cap N$ means. My initial thoughts are that a nonzero vector is a vector such that $x_1, x_2, x_3 \neq 0$, but not sure where to go from here
 A: A nonzero vector means that at least one of $x_1,x_2,x_3$ is not zero. The zero vector has all components equal to $0$. You simply need to solve the system of equations
$$x_1-x_2+2x_3=0$$
$$x_1+x_2+x_3=0$$
 to find such a vector. This will also help you show that all solutions are scalar multiples. 
A: A nonzero vector is simply a vector that is not the zero vector. That is, it is not the case that $x_1 = x_2 = x_3 = 0$. Notice that this is not the same as $x_1, x_2, x_3 \neq 0$, as, for example, the vector $(1,0,0)$ is nonzero.
The set $M \cap N$ is the set of triples $(x_1, x_2, x_3)$ such that both of the following happen:
$x_1 - x_2 + 2 x_3 = 0$ (that is, the vector is in $M$)
$x_1 + x_2 + x_3 = 0$ (that is, the vector is in $N$)
This is a system of two equations that can easily be shown to be equivalent to the system:
$x_1 + \frac 3 2 x_3 = 0$
$x_1 + 3 x_2 = 0$
Now the following fact becomes obvious: a vector from the space $M \cap N$ is uniquely determined by its first component. That is, knowing its first component we can determine the other two, and for any $x_1$ we can find a vector of $M \cap N$ such that its first component is $x_1$.
Taking this a step further, we come to the conclusion (solving the previous equations for $x_2$ and $x_3$) that the vectors of $M \cap N$ are precisely those of the form:
$(x_1, -\frac 1 3 x_1, - \frac 2 3 x_1)$, or, equivalently, $x_1 (1, -\frac 1 3, -\frac 2 3)$. The interpretation of this is that, if we call $a = (1, \, -\frac 1 3, -\frac 2 3)$, we have $M \cap N$ is the set of vectors of the form $k a$, for $k \in \mathbb{R}$, which is precisely what it means for it to be generated by $a$.
Edit: As requested, clarification for the system equivalence.
The method I used was (not quite) gaussian elimination. I say not quite because gaussian elimination would have gotten me the results as a function of $x_3$ and not $x_1$. Regardless, I will outline the method of (proper) gaussian elimination, and leave it to the reader to revise the above proof to work with $x_3$. One should end up with a different vector than above, and I encourage the reader to interpret why both vectors work.
The system we begin with is as follows:
$x_1 - x_2 + 2 x_3 = 0$
$x_1 + x_2 + x_3 = 0$
We wish to get rid of the first variable in the second line. As such, we subtract the first line from the second, yielding
$x_1 - x_2 + 2 x_3 = 0$
$2 x_2 - x_3 = 0$
Now, to get rid of the second variable in the first equation, we add half the second equation to the first. This gets us to
$x_1 + \frac 3 2 x_3 = 0$
$2 x_2 - x_3 = 0$
From this, one may write $x_1$ and $x_2$ as a function of $x_3$, and the rest follows.
