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If $$\Delta =\begin{vmatrix} \arg\ { z }_{ 1 } & \arg\ { z }_{ 2 } & \arg\ { z }_{ 3 } \\ \arg\ { z }_{ 2 } & \arg\ { z }_{ 3 } & \arg\ { z }_{ 1 } \\ \arg\ { z }_{ 3 } & \arg\ { z }_{ 1 } & \arg\ { z }_{ 2 } \end{vmatrix}$$
then $$\Delta$$ is divisible by


A
arg (z1+z2+z3)
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B
argz1z2z3
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C
argz1+argz2+argz3
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D
none
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Solution

The correct options are
B $$\arg{z}_{1}{z}_{2}{z}_{3}$$
D $$\arg{z}_{1}+\arg{z}_{2}+\arg{z}_{3}$$
Ans. $$(b).\ (c).$$
Apply $${C}_{1}+{C}_{2}+{C}_{3}$$ and take $$\displaystyle \sum \arg\ {z}_{1}$$ common.
$$\therefore \Delta = \sum { \left( \arg{ z }_{ 1 } \right)  } \begin{vmatrix} 1 & \arg{ z }_{ 2 } & \arg{ z }_{ 3 } \\ 1 & \arg{ z }_{ 3 } & \arg{ z }_{ 1 } \\ 1 & \arg{ z }_{ 1 } & \arg{ z }_{ 2 } \end{vmatrix}$$
Hence $$\Delta$$ is divisible by $$\sum \arg\  {z}_{1}$$
$$\arg{ z }_{ 1 }+\arg{ z }_{ 2 }+\arg{ z }_{ 3 }=\arg\ \left( { z }_{ 1 }\ { z }_{ 2 }\ { z }_{ 3 } \right).$$

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