Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&=3\\
x+a^2y+4z&=3+a
\end{align*}
\end{cases}
$$
and solve for x, y and z. I subtract the first row from the second and third
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(a^2-1)y+3z&=1+a.
\end{align*}
\end{cases}
$$
I subtract $(a+1)*\textrm{row}_1$ from the third row:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(2-a)z&=1+a.
\end{align*}
\end{cases}
$$
I think I am supposed to find an $a$ such that $0=0$ in the third row, but obviously there isn't any such $a$. Do I have the right idea about how to solve this? If so, where is my mistake?

You had the right idea. 
Of yourse, in this case, there was only one parameter, and even that appeared in the form of simple expressions ($a$ and $a^2$). With more complicated problems, I suggest you start in a different way. 
Compute the determinant of the matrix of coeeficients on the LHS: 
$$\det\begin{pmatrix} 1 & 1 & 1 \\ 1 & a & 2 \\ 1 & a^2 & 4 \\  \end{pmatrix} = 4a+2+a^2-2a^2-4-a= -a^2+3a-2$$
This is zero iff $a=2$ or $a=1$. These are the only values to check: if the determinant is nonzero, then the system has a unique solution, so the planes intersect in a single point.
