If $x + y + z = 0$, then prove that $\left| {\begin{array}{llllllllllllllllllll}{xa}&{yb}&{zc}\\{yc}&{za}&{xb}\\{zb}&{xc}&{ya}\end{array}} \right| = xyz\left| {\begin{array}{cccccccccccccccccccc}a&b&c\\c&a&b\\b&c&a\end{array}} \right|$.

Since, $x + y + z = 0$, also we have to prove $\left| {\begin{array}{llllllllllllllllllll}{xa}&{yb}&{zc}\\{yc}&{za}&{xb}\\{zb}&{xc}&{ya}\end{array}} \right| = xyz\left| {\begin{array}{llllllllllllllllllll}a&b&c\\c&a&b\\b&c&a\end{array}} \right|$

$\therefore$ ${\rm{LHS}} = \left| {\begin{array}{llllllllllllllllllll}{xa}&{yb}&{zc}\\{yc}&{za}&{xb}\\{zb}&{xc}&{ya}\end{array}} \right|$

$= xa(za \cdot ya - xb \cdot xc) - yb(yc \cdot ya - xb \cdot zb) + zc(yc \cdot xc - za \cdot zb)$

$= xa\left( {{a^2}yz - {x^2}bc} \right) - yb\left( {{y^2}ac - {b^2}xz} \right) + zc\left( {{c^2}xy - {z^2}ab} \right)$

$= xyz{a^3} - {x^3}abc - {y^3}abc + {b^3}xyz + {c^3}xyz - {z^3}abc$

$= xyz\left( {{a^3} + {b^3} + {c^3}} \right) - abc\left( {{x^3} + {y^3} + {z^3}} \right)$
$= xyz\left( {{a^3} + {b^3} + {c^3}} \right) - abc(3xyz)$

$= xyz\left( {{a^3} + {b^3} + {c^3} - 3abc} \right)$

…….(i)
Now, ${\rm{RHS}} = xyz\left| {\begin{array}{llllllllllllllllllll}a&b&c\\c&a&b\\b&c&a\end{array}} \right| = xyz\left| {\begin{array}{llllllllllllllllllll}{a + b + c}&b&c\\{a + b + c}&a&b\\{a + b + c}&c&a\end{array}} \right|$

$= xyz(a + b + c)\left| {\begin{array}{cccccccccccccccccccc}1&b&c\\1&a&b\\1&c&a\end{array}} \right|$

[taking $(a + b + c)$ common from ${C_1}$]
$= xyz(a + b + c)\left| {\begin{array}{cccccccccccccccccccc}0&{b - c}&{c - a}\\0&{a - c}&{b - a}\\1&c&a\end{array}} \right|$

and $\left. {{R_2} \to {R_2} - {R_3}} \right]$

Expanding along ${C_1}$,

$= xyz(a + b + c)[1(b - c)(b - a) - (a - c)(c - a)]$

$= xyz(a + b + c)\left( {{b^2} - ab - bc + ac + {a^2} + {c^2} - 2ac} \right)$

$= xyz(a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$

$= xyz\left( {{a^3} + {b^3} + {c^3} - 3abc} \right)$

………(ii)
From Eqs. (i) and (ii),
${\rm{LHS}} = {\rm{RHS}}$

$\Rightarrow$ $\left| {\begin{array}{llllllllllllllllllll}{xa}&{yb}&{zc}\\{yc}&{za}&{xb}\\{zb}&{xc}&{ya}\end{array}} \right| = xyz\left| {\begin{array}{llllllllllllllllllll}a&b&c\\c&a&b\\b&c&a\end{array}} \right|$