Geometric Interpretation of System of Equations Perhaps I am overlooking something but this is puzzling me. 
I have a system of equations which can be represented by three planes:
\begin{align}
\Pi_1&: \ x-7y+5z=3\\
\Pi_2&: \ x-y+3z=-1\\
\Pi_3&: \ 3y-z=k
\end{align}
I have made an augmented matrix in echelon form:
$$ \left[
\begin{array}{ccc|c}
  1&-7&5&3\\
  0&-6&2&4\\
  0&0&0&2k+4\\
\end{array}
\right] $$
Now consider when this system of equations has infinite solutions. It should be when $2k+4=0\implies k=-2$, right? Then either $0x=0$ or $0y=0$ or $0z=0$ which is always true.
Furthermore notice that $\Pi_3$ is actually a linear combination of $\Pi_1$ and $\Pi_2$ (when $k=-2$) because $\Pi_3=\frac{1}{2}\Pi_2 - \frac{1}{2}\Pi_1$. Doesn't this mean that $\Pi_3$ passes through the solution set of $\Pi_1$ and $\Pi_2$?
If the solution set is indeed infinite, the intersection between these three planes should be one line. However when I check on GeoGebra and it shows this:

which means I am incorrect. What am I missing here? Thanks for the help in advance.
 A: The three planes do indeed intersect in a line when $k=-2$. Check what it is that you actually graphed in GeoGebra. It looks to me like the third plane in your image is $3x-y=-2$ instead of $3y-z=-2$. The latter is not orthogonal to the $x$-$y$ plane as the plane in your image is.  

We can verify this intersection by using the Plücker matrix of the intersection line of the first two planes. One parameterization of this line is $\frac13(-5,-2,0)+\lambda(-8,1,3)$. In homogeneous coordinates, then, this line is the join of $\mathbf p = (-5:-2:0:3)$ and $\mathbf q=(-8:1:3:0)$. If we treat these as column vectors, the Plücker matrix of this line is $\mathcal L = \mathbf p\mathbf q^T-\mathbf q\mathbf p^T$. The intersection of this line with a plane $\mathbf\pi$ is then given by $$\mathcal L\mathbf\pi = (\mathbf p\mathbf q^T-\mathbf q\mathbf p^T)\mathbf\pi = (\mathbf\pi^T\mathbf q)\mathbf p - (\mathbf\pi^T\mathbf p)\mathbf q.$$ This vanishes iff the line represented by $\mathcal L$ lies on $\mathbf\pi$. For the third plane, $\mathbf\pi = (0:3:-1:-k)$, for which $$\mathcal L\mathbf\pi = (k+2)\begin{bmatrix}-24\\3\\9\\0\end{bmatrix},$$ which vanishes when $k=-2$ as you determined by other means.
A: Hint: Multiplying your first equation by $$-1$$ and adding to the second we get
$$-8y-2z=-4$$ multiplying the last equation by $$-2$$ and adding to the equation above we obtain
$$-14y=-4-2k$$
