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(D) First,find the point of intersection of $L_1: 2x + 3y + 6 = 0$ and $L_2: 3x - y - 13 = 0$.Multiply $L_2$ by $3$: $9x - 3y - 39 = 0$.Adding this to

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$3x - 4y - 25 = 0$

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(D) First,find the point of intersection of $L_1: 2x + 3y + 6 = 0$ and $L_2: 3x - y - 13 = 0$.Multiply $L_2$ by $3$: $9x - 3y - 39 = 0$.Adding this to $L_1$: $(2x + 3y + 6) + (9x - 3y - 39) = 0$ $\Rightarrow 11x - 33 = 0$ $\Rightarrow x = 3$.Substitute $x = 3$ into $L_2$: $3(3) - y - 13 = 0$ $\Rightarrow 9 - y - 13 = 0$ $\Rightarrow y = -4$.The point of intersection is $(3, -4)$.Any line parallel to $3x - 4y + 5 = 0$ is of the form $3x - 4y + k = 0$.Since this line passes through $(3, -4)$,substitute the coordinates:$3(3) - 4(-4) + k = 0$ $\Rightarrow 9 + 16 + k = 0$ $\Rightarrow 25 + k = 0$ $\Rightarrow k = -25$.Thus,the required equation is $3x - 4y - 25 = 0$.

The equation of the line passing through the point of intersection of the lines $2x + 3y + 6 = 0$ and $3x - y - 13 = 0$ and parallel to the line $3x - 4y + 5 = 0$ is

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