Is there anything wrong with this question? Find an equation of the plane that passes thru the points $p(1,0,1)$ and $q(2,1,0)$ which is parallel to the intersection of the two planes $x+y+z=5$ and $3x-y=4$?
I plotted the two points and the intersection of the two planes using a 3-d grapher and the points aren't parallel which means the equation of the plane requested wouldn't be parallel? Is that correct?
 A: I believe what's being asked is to find a plane containing the two points $p$ and $q$, such that the plane never hits the line of intersection of the two planes $x+y+z=5$ and $3x-y=4$. 
If you visualize this, you have a pair of lines $L_1$ (connecting $p$ and $q$) and $L_2$ (the intersection of the other two planes); as long as $L_1$ and $L_2$ themselves don't intersect, we can find a plane containing $L_1$ not meeting $L_2$. Note that this doesn't require $L_1$ and $L_2$ to be parallel, only skew (=non-intersecting). Think about why!
A: It's the plane that needs to be parallel, not the points.
The points just need to lie in the plane.
Let the equation of the plane be $ax+by+cz=d$. Without loss of generality we can let $d=1$.
Using the values of $(x,y,z)$ from the given points, we find that $a+c=d=1$ and that $2a+b=d=1$.
The line of intersection has direction vector equal to the vector product of the normals to the two planes:
$\underline p=\left(\begin{array}{r}
1 \\
1 \\
1
\end{array}\right) \times
\left(\begin{array}{r}
3 \\
-1 \\
0
\end{array}\right)=\left(\begin{array}{r}
1 \\
3 \\
-4
\end{array}\right)
$
This direction vector is parallel to the required plane, so 
$\left(\begin{array}{r}
1 \\
3 \\
-4
\end{array}\right) .
\left(\begin{array}{r}
a \\
b \\
c
\end{array}\right)=0
$
$a+3b-4c=0$
Now pull together these three equations:
$a+3b-4c=0$
$a+c=1$
$2a+b=1$
Let $c=1-a$, so that our equations are:
$a+3b-4+4a=0 \Rightarrow 5a+7b=4$
$2a+b=1 \Rightarrow 14a+7b=7$
These give $9a=3 \Rightarrow a=\frac 1 3$, $b= \frac 1 3$, $c=\frac 23$
Plane is $\frac13x+\frac13y+\frac23z=1$ or better $x+y+2z=3$.
