Taking $\left(b - a\right)$ common from $C_{1}$ and $C_{3}$, we get
$\Delta=\left(b-a\right)^{2}\left|\begin{matrix}b&b-c&c\\ a&a-b&b\\ c&c-a&a\end{matrix}\right|$
Applying $C_{1} \rightarrow C_{1}- C_{2}$, we get
$\Delta=\left(b-a\right)^{2} \left|\begin{matrix}c&b-c&c\\ b&a-b&b\\ a&c-a&a\end{matrix}\right|$
Since, $C_{1}$ and $C_{3}$ are identical
$\Rightarrow \quad\Delta=0$