If {z_1}, {z_2}, {z_3} are three non-zero com | vedclass

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(C) Given the determinant $\left| {\begin{array}{*{20}{c}} a & b & c \ b & c & a \ c & a & b \end{array}} ight| = 0$. Expanding the determinant,we

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$arg{\left( {\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}}} \right)^2}$

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(C) Given the determinant $\left| {\begin{array}{*{20}{c}} a & b & c \ b & c & a \ c & a & b \end{array}} ight| = 0$. Expanding the determinant,we get $-(a^3 + b^3 + c^3 - 3abc) = 0$,which implies $(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$. Since $a, b, c$ are moduli of non-zero complex numbers,$a, b, c > 0$,so $a+b+c 
eq 0$. Thus,$a^2+b^2+c^2-ab-bc-ca = 0$,which simplifies to $\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$. This implies $a=b=c$. Given the geometric properties of complex numbers and the provided options,the expression simplifies to $arg{\left( {\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}}} ight)^2}$.

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