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(D) Let $\Delta = \left| \begin{array}{ccc} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{array} ight|$.Applying the row operation $R_1 	o R_1 + R

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$(0, 0)$

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(D) Let $\Delta = \left| \begin{array}{ccc} 6i & -3i & 1 \ 4 & 3i & -1 \ 20 & 3 & i \end{array} ight|$.Applying the row operation $R_1 	o R_1 + R_2$:$\Delta = \left| \begin{array}{ccc} 6i+4 & 0 & 0 \ 4 & 3i & -1 \ 20 & 3 & i \end{array} ight|$.Expanding along the first row:$\Delta = (6i + 4) 	imes \left| \begin{array}{cc} 3i & -1 \ 3 & i \end{array} ight| - 0 + 0$.$\Delta = (6i + 4) 	imes ((3i 	imes i) - (-1 	imes 3))$.$\Delta = (6i + 4) 	imes (3i^2 + 3)$.Since $i^2 = -1$,we have $3i^2 + 3 = 3(-1) + 3 = -3 + 3 = 0$.Therefore,$\Delta = (6i + 4) 	imes 0 = 0$.Given $\Delta = x + iy$,we have $x + iy = 0 + i(0)$.Comparing real and imaginary parts,we get $(x, y) = (0, 0)$.

If $\left| \begin{array}{ccc} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{array} \right| = x + iy$,then $(x, y)$ is

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