Intersection between three planes - geometric configurations for non-unique solutions

Question

I have three planes with equations

$$x+y+kz=-2$$ $$3x-y+14z=6$$ $$x+ky = -5$$There is no unique solution for $k=2$ and $k=\frac{7}{3}$, since that is when the determinant of the matrix of coefficients is $0$.

I want to look at the geometric configurations of the three planes for these values of $k$. For $k=2$, it is simple, as they form a sheaf and meet in line.

However, I am stuck with the case $k=\frac{7}{3}$. Do they form a prism or some other configuration? I am not sure on the conditions for them to form a prism. I have tried to plot these on a 3D plotter for this case, but I still can't seem to see if they form a prism or some other configuration.

Any help is appreciated.

Answer 1

There is no solution when $k=7/3$. Even if there was a solution you can only get a line or a plane in the case when the determinant is zero.

Answer 2

Let’s look at it from a projective-geometric point of view. The intersection of the three planes is the null space of the matrix $$\begin{bmatrix}1&1&k&2\\3&-1&14&-6\\1&k&0&5\end{bmatrix}.$$ When $k=2$ this space is two dimensional, with basis $\{[1,-3,0,1]^T,[-4,2,1,0]^T\}$, i.e., the line $(1,-3,0)+\lambda(-4,2,1)$. Otherwise, the null space is one-dimensional and is spanned by $[6k+35,-21,-12,3k-7]^T$. When $k=\frac73$, this is the point at infinity $[49,-21,-12,0]^T$, which indicates that the three pairwise intersections are parallel lines in the direction of $(49,-21,-12)$. To find the specific lines, you’ll have to compute those pairwise intersections.