By using properties of determinants,show that:
$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

(A) Let $\Delta = \left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$.
This is a skew-symmetric determinant of odd order $3 \times 3$.
We can expand the determinant along the first row:
$\Delta = 0(0 - (-c^2)) - a(0 - (-bc)) + (-b)(-ac - 0)$
$\Delta = 0 - a(bc) - b(-ac)$
$\Delta = -abc + abc$
$\Delta = 0$.
Alternatively,by property of determinants,if we take $-1$ common from $R_1, R_2,$ and $R_3$,we get:
$\Delta = (-1)^3 \left|\begin{array}{ccc}0 & -a & b \\ a & 0 & c \\ -b & -c & 0\end{array}\right| = -\Delta^T$
Since $\Delta = -\Delta$,we have $2\Delta = 0$,which implies $\Delta = 0$.