By using properties of determinants, show that:

We have,

$\Delta=\left|\begin{array}{lll}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$

Applying $\mathrm{R}_{1} \rightarrow c \mathrm{R}_{1}$, we have:]

$\Delta=\frac{1}{c}\left|\begin{array}{lll}0 & a c & -b c \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$

Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-b \mathrm{R}_{2}$, we have:

$\begin{aligned} \Delta &=\frac{1}{c}\left|\begin{array}{lll}a b & a c & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \\ &=\frac{a}{c}\left|\begin{array}{lll}b & c & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \end{aligned}$

Here, the two rows R1 and R3 are identical.

$\therefore \Delta=0$.