If $\Delta =\left|\begin{array}{ccc}0& a-b& a-c\\ b-a& 0& b-c\\ c-a& c-b& 0\end{array}\right|$, then $\Delta$ equals:

The given determinant is:
$$\Delta =\left|\begin{array}{ccc}0& a-b& a-c\\ b-a& 0& b-c\\ c-a& c-b& 0\end{array}\right|$$
Notice that this is a skew-symmetric matrix $M$ where $M^T=-M$.For any skew-symmetric matrix $M$ of odd order (here $3\times 3$), the determinant is always $0$.Alternatively, applying $C_1\to C_1-C_3$ and $C_2\to C_2-C_3$:
$$\Delta =\left|\begin{array}{ccc}-(a-c)& -(b-c)& a-c\\ -(a-c)& -(b-c)& b-c\\ -(a-c)& -(b-c)& 0\end{array}\right|$$
Taking $-(a-c)$ common from $C_1$ and $-(b-c)$ common from $C_2$:
$$\Delta =(a-c)(b-c)\left|\begin{array}{ccc}1& 1& a-c\\ 1& 1& b-c\\ 1& 1& 0\end{array}\right|$$
Since columns $C_1$ and $C_2$ are identical, the value of the determinant is $0$.