Consider the following linear equations $ax + by + cz = 0,bx + cy + az = 0,cx + ay + bz = 0$

Here we have the determinant of the coefficient

$\text{ matrix of given equation as }\mathrm{\Delta }=\left|\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\end{array}\right|$

$=-(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$

$=-\frac{1}{2}(a+b+c)\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]$

$\text{ A) }a+b+c\ne 0\text{ and }{a}^2+b^2+c^2-ab-bc-ca=0$

$\text{ or }(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0\text{ or }a=b=c$

Therefore, this equation represents identical planes.

$\text{ B) }a+b+c=0\text{ and }{a}^2+b^2+c^2-ab-bc-ca\ne 0$

This means  $\mathrm{\Delta }=0$ and a, b and c are not all equal.

Therefore, all equations are not identical but have infinite solutions.

Hence, $ax + by = (a + b)z$(using $a+b+c=0$) and $bx + cy$
$= (b + c)z$

$\Rightarrow \left(b^2-ac\right)y=\left(b^2-ac\right)z\Rightarrow y=z$

$\Rightarrow ax+by+cy=0\Rightarrow ax=ay$

$\Rightarrow x=y=z$

Therefore, the equations represent the line$x = y = z$

$\text{ C) }a+b+c\ne 0\text{ and }{a}^2+b^2+c^2-ab-bc-ca=0$

$\Rightarrow \mathrm{\Delta }\ne 0$ and the equations have only trivial solution, i.e., $x = y = z = 0$

Therefore, the equation represent the planes meeting at a single point, namely origin.

$\text{ D) }a+b+c\ne 0\text{ and }{a}^2+b^2+c^2-ab-bc-ca=0$ 

$\Rightarrow a=b=c\text{ and }\mathrm{\Delta }=0\Rightarrow a=b=c=0$

all equations are satisfied by all x, y and z The equations represent the whole of the threedimensional space.