Finding a value a so that 3 planes don't intersect in a single point I'm currently stuck at the following question:
Three planes have equation:
$$ax+2y+z=3$$
$$-x+(a+1)y+3z=1$$
$$-2x+y+(a+2)z=k$$
where $a∈R$.
Question: Find a value $a$ for which these planes do not intersect in a point.
The easiest way would be to make one equation a multiple of another equation but this isn't possible at first glance. Then I tried to eliminate $x$ from equation $1$ and $2$:
\begin{bmatrix}a&2&1&3\\-1&a+1&3&1\\-2&1&a+2&k\end{bmatrix}
\begin{bmatrix}a&2&1&3\\2&-2a-1&-6&-2\\-2&1&a+2&k\end{bmatrix}
\begin{bmatrix}a&2&1&3\\0&-2a&a-4&k-2\\-2&1&a+2&k\end{bmatrix}
However, this did not get me very far.
The correct solution is said to be $a = 1$ but I don't understand why. Is there a generell procedure to make a system of equations have one, infinitely many or no solutions?

Edit: After having found $a = 1$, by using gaussian elimination I found a value for $k$ such that the lines intersect in a line:
\begin{bmatrix}1&2&1&3\\-1&2&3&1\\-2&1&3&k\end{bmatrix}
\begin{bmatrix}1&2&1&3\\0&4&4&4\\-2&1&3&k\end{bmatrix}
\begin{bmatrix}2&4&2&6\\0&4&4&4\\-2&1&3&k\end{bmatrix}
\begin{bmatrix}2&4&2&6\\0&4&4&4\\0&5&5&k+6\end{bmatrix}
$$∴ k+6 = 5$$
$$ k = -1 $$
If $a=1$,$k=-1$ the planes therefore intersect at:
$$4y+4z=4$$
$$y+z=1$$
$$let\  z = \lambda $$
$$y = 1- \lambda$$
$$2x + 4y + 2z = 6$$
$$x = 1 + \lambda$$
$$l:\vec r = \begin{bmatrix}1\\1\\0\end{bmatrix} + \lambda \begin{bmatrix}1\\-1\\1\end{bmatrix}$$
For those struggling with a similar question :D
 A: Just calculate the determinant that those vectors define:
$$\begin{vmatrix}
a & 2 & 1 \\
-1 & a+1 & 3 \\
-2 & 1 & a+2
\end{vmatrix} = a^3+3a^2+3a-7$$
Then, the values of $a$ that will lead to the planes not intersecting at a single point will be the ones that verify
$$a^3+3a^2+3a-7=0,$$
and that just gives the value $a=1$ (considering $a\in\mathbb{R}$, since the other two roots of the polynomial are complex). The explanation to this is that if the determinant is $0$, then those vectors are not linearly independent, so the space they generate isn't of dimension $3$ and we conclude that the planes will intersect either nowhere, either at one plane, or either at one, two or three lines, so that makes it impossible for them to intersect at just one single point.
A cool interpretation of this problem is that your system of equations is indeed the representation of three different planes, and the solution to it will be the intersection between those planes. But you must notice that your system has one single solution if and only if the determinant of the system is different from $0$, so it will have either none or infinite solutions if it's equal to $0$, the case when $a=1$.
