What does the homogeneous system of equations represent under certain conditions? 
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.
 A: 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.
A: 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).
