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?

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 $x+y+2z=3$.
