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$If x + iy = |\begin{matrix} 6 i & - 3 i & 1 \\ 4 & 3 i & - 1 \\ 20 & 3 & i \end{matrix}| then$

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$x = 3, y = 1$

$x = 1, y = 3$

$x = 0, y = 3$

$x = 0, y = 0$

answer is D.

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Detailed Solution

$\begin{array}{rcl} x + iy & = & 6 i (- 3 + 3) + 3 i (4 i + 20) + 12 - 60 i \\ = & 0 - 12 + 60 i + 12 - 60 i \\ = & 0 \\ x + iy & = & 0 + i 0 \\ ∴ x & = & 0, y = 0 \end{array}$

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