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(D) Let the given determinant be $\Delta = \left| \begin{array}{ccc} 1 & x & y \ 2 & \sin x + 2x & \sin y + 2y \ 3 & \cos x + 3x & \cos y + 3y \end{

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$\sin(x - y)$

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(D) Let the given determinant be $\Delta = \left| \begin{array}{ccc} 1 & x & y \ 2 & \sin x + 2x & \sin y + 2y \ 3 & \cos x + 3x & \cos y + 3y \end{array} ight|$.Applying row operations $R_{2} ightarrow R_{2} - 2R_{1}$ and $R_{3} ightarrow R_{3} - 3R_{1}$:$\Delta = \left| \begin{array}{ccc} 1 & x & y \ 2 - 2(1) & (\sin x + 2x) - 2x & (\sin y + 2y) - 2y \ 3 - 3(1) & (\cos x + 3x) - 3x & (\cos y + 3y) - 3y \end{array} ight|$$\Delta = \left| \begin{array}{ccc} 1 & x & y \ 0 & \sin x & \sin y \ 0 & \cos x & \cos y \end{array} ight|$Expanding along the first column:$\Delta = 1 \cdot (\sin x \cos y - \cos x \sin y)$Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:$\Delta = \sin(x - y)$.

The value of $\left| \begin{array}{ccc} 1 & x & y \\ 2 & \sin x + 2x & \sin y + 2y \\ 3 & \cos x + 3x & \cos y + 3y \end{array} \right|$ is

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