If $p+q+r=0$, prove that
$$\begin{vmatrix} pa&qb&rc\\ qc&ra&pb\\ rb&pc&qa \end{vmatrix}=pqr\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a \end{vmatrix}$$
My attempt is as follows:-
Taking $p$ common from first row, $q$ from second row and $r$ from third row
$$pqr\begin{vmatrix} a&\dfrac{qb}{p}&\dfrac{rc}{p}\\ c&\dfrac{ra}{q}&\dfrac{pb}{q}\\ b&\dfrac{pc}{r}&\dfrac{qa}{r} \end{vmatrix}$$
$$pqr\begin{vmatrix} a&\dfrac{-bp-br}{p}&\dfrac{-cp-cq}{p}\\ c&\dfrac{-pa-qa}{q}&\dfrac{-qb-rb}{q}\\ b&\dfrac{-qc-rc}{r}&\dfrac{-pa-ra}{r} \end{vmatrix}$$
$$pqr\begin{vmatrix} a&-b-\dfrac{br}{p}&-c-\dfrac{cq}{p}\\ c&-a-\dfrac{pa}{q}&-b-\dfrac{rb}{q}\\ b&-c-\dfrac{qc}{r}&-a-\dfrac{pa}{r} \end{vmatrix}$$
$$pqr\begin{vmatrix} a&b+\dfrac{br}{p}&c+\dfrac{cq}{p}\\ c&a+\dfrac{pa}{q}&b+\dfrac{rb}{q}\\ b&c+\dfrac{qc}{r}&a+\dfrac{pa}{r} \end{vmatrix}$$
$$pqr\left(\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\end{vmatrix}+ \begin{vmatrix} a&\dfrac{br}{p}&c\\ c&\dfrac{pa}{q}&b\\ b&\dfrac{qc}{r}&a\end{vmatrix}+\begin{vmatrix} a&b&\dfrac{cq}{p}\\ c&a&\dfrac{rb}{q}\\ b&c&\dfrac{pa}{r} \end{vmatrix} \right) $$
$$\left(pqr\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\end{vmatrix}+ \begin{vmatrix} pa&br&pc\\ qc&pa&qb\\ rb&qc&ra\end{vmatrix}+\begin{vmatrix} pa&pb&cq\\ qc&qa&rb\\ rb&rc&pa \end{vmatrix} \right) $$
Now I am left with two other determinants and their sum should come out zero.
I am getting no way to simplify those determinants, if you open those determinants, expression would be quite large.
Is there any better way to solve this problem?