Given system of equations has a non-trivial solution
$\begin{aligned} & \Pi_1 \equiv x+a y+a z=0 \\ & \Pi_2 \equiv b x+y+b z=0 \\ & \Pi_3 \equiv c x+c y+z=0\end{aligned}$
It can be written as $A X=0$
Where, $\quad A=\left[\begin{array}{lll}1 & a & a \\ b & 1 & b \\ c & c & 1\end{array}\right]$
So, the system has non-trivial solutions hence it has infinitely many solutions or unique solutions.
$\therefore \quad|A|=0$
Now, let the system has infinitely many solutions So, let $y=t ; t \in \mathbf{R}$
Then from Eqs. (i) and (ii), we get
$x+a z=-a t \Rightarrow b x+b z=-t$
Solving these, we get
$z=\frac{t(1-a b)}{b(a-1)}, x=\frac{a}{b} t \frac{(1-b)}{(1-a)}$
Hence, if $a \neq b \neq c$, then $x=\frac{a}{b} t \frac{(1-b)}{(1-a)}, y=t$,
$z=\frac{t}{b} \frac{(1-a b)}{(a-1)}, t \in \mathbf{R}$
If $a=b=c$
Let $y=t_1, z=t_2, t_1, t_2 \in \mathbf{R}$
Then by Eq. (i)
$x=-a\left(t_1+t_2\right)$
So, the system of equations has unique solution.