Let $\mathrm{\Delta }=\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2-bc\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}^2-ca\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}^2-ab\end{array}\right|$ , then$∆$ is equal to

 

Applying $R_2\to R_2-R_1+R_3\to R_3-R_1$ we get

$\mathrm{\Delta }=(b-a)(c-a)\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2-bc\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+b+c\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+b+c\end{array}\right|=0$

                              $\left[\because R_2\text{ and }{R}_3\text{ are identical }\right]$