Using the property of determinants and without expanding, prove that:

$\Delta=\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}$, we have:

$\Delta=\left|\begin{array}{lll}a-c & b-a & c-b \\ b-c & c-a & a-b \\ -(a-c) & -(b-a) & -(c-b)\end{array}\right|$

$=-\left|\begin{array}{ccc}a-c & b-a & c-b \\ b-c & c-a & a-b \\ a-c & b-a & c-b\end{array}\right|$

Here, the two rows R1 and R3 are identical.

$\therefore \Delta=0 .$