Question 1048128
X+Y+Z= -1
2X-Y+2Z=-5
-X+2Y-Z=4
first and third
x+y+z=-1
-x+2y-z=4
3y=3
y=1
Therefore, x+1+z=-1
x+z=-2
substitute into third x=-z-2
z+2+2-z=4
4=4
-z-2,1,z are solutions
if z=0
-2+1=-1
if z=8
-10+1+8=-1
They aren't independent
(-z-2,1,z} are the points.  Infinite solutions.
The determinant of the x,y,z matrix is 0, so there is linear dependency. One variable is a linear function of another (x and z).