many thanks for your help, I think that’s the solution.

Chris

P.S.

By the way, here is a small typo
https://www.cut-the-knot.org/WhatIs/WhatIsPoint.shtml
“…without width, breath (!) or length.” (third line)

5, 6 ,7
8, 9, 10 =0
11, 12, 13

More general, if determinant 3x3 contains of three arithmetic series, its value is zero.
The middle one being the average of the other two. (Alex)

This is very useful for teaching, you can create determinant with any value. For example, create determinant 2x2 with value 1:

8 3
5 2 =1 (8*2-5*3)

lets extend this to be arithmetic :

8 3 -2
5 2 -1 = 0 (see above)
a a+d a+2d

if we add number x to right - down corner (a+2d+x) determinant will have value x. For example:

8 3 -2
5 2 -1 =
-4 -2 x

8 3 -2
5 2 -1 +
-4 -2 0

8 3 -2
5 2 -1 = 0 + x*(8*2-5*3) = x
0 0 x

we are using rule 3 from the article "What Is Determinant".
It is powerful tool if we want to write by heart on the blackboard determinant for the practicing. I hope it is useful for someone.

It is nice to be here again!!!

| 8  3  -2|   | 8  3 -2|   | 8  3 -2|
| 5  2  -1| = | 5  2 -1| + | 5  2 -1|
|-4 -2   x|   |-4 -2  0|   | 0  0  x|

is indeed very useful.

I am grateful you found the time.

| 8 3 -2|
| 5 2 -1| =
| 0 0 x|

|3 -2| * 0 -
|2 -1|

|8 -2| * 0 +
|5 -1|

|8 3| * x =1 * x = x
|5 2|

Second example, value is even:

|-1 -5| = 2
|-1 -7|

extend that rows to the left to be arithmetic:

|3 -1 -5|
|5 -1 -7|

add another arithmetic row |-1 -4 -7| to the bottom (or top)

|3 -1 -5|
|5 -1 -7| = 0
|-1 -4 -7|

and finally:

|3 -1 -5|
|5 -1 -7| = 2x
|x-1 -4 -7|

Only because you have not stated this explicitly.