What does the homogeneous system of equations represent under certain conditions?

Question

Consider the following linear equations

$ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$

1) $a+b+c \neq o$ and $a^2+b^2+c^2=ab+bc+ca$

2) $a+b+c \neq o$ and $a^2+b^2+c^2 \neq ab+bc+ca$

3) $a+b+c = o$ and $a^2+b^2+c^2 \neq ab+bc+ca$

4) $a+b+c = o$ and $a^2+b^2+c^2 = ab+bc+ca$

Now I need to match these with the following options.

a) The equations represent planes meeting only at a single point.

b) The equations represent the line $x=y=z$

c) The equations represent identical planes

d) The equations represent the whole 3D space.

I found out the determinant using Cramer's rule i.e. $-(a^3+b^3+c^3-3abc)$.So using Cramer's rule whenever the determinant is 0 there should be infinite solutions.After that I'm confused as to how to proceed.

Answer 1

The determinant is : $$\begin{align}\begin{vmatrix} a&b&c\\\ b&c&a \\\ c&a&b \end{vmatrix}&=3abc-a^3-b^3-c^3 \\ &=-\frac 12(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right) \\ &=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \end{align} $$

1)This one implies $a=b=c\ne0$ so they are identical planes.

2)Here the determinant is non-zero so these are planes meeting at single point.

3) Here the determinant is zero so there are infinite solutions. And if you put any point $(x,x,x)$ in the equation it will always satisfy, so the intersection of these planes is the line $x=y=z$

4)Here $a=b=c=0$ so any point in the space is satisifed.

Answer 2

hint

Note that $a^2+b^2+c^2=ab+bc+ca$ is another way of saying $a=b=c$ (just multiply both sides by $2$ and complete the squares).

So if (4) holds, then $a=b=c=0$, in which case the system is actually $0=0$, hence the solution space is all of $\mathbb{R}^3$.

Try to proceed and see if you can complete now. Otherwise I can elaborate.

Further elaboration:

Consider (1), since $a+b+c \neq 0$, then we know even though $a=b=c$ but they are not zero. In which case we are essentially given one plane, namely $x+y+z=0$ That corresponds to (c).