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(C) First,find the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$.Multiplying the second equation by $3$,we get $9x + 6y +

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$5x + 3y + 8 = 0$

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(C) First,find the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$.Multiplying the second equation by $3$,we get $9x + 6y + 15 = 0$.Adding this to the first equation: $(5x - 6y - 1) + (9x + 6y + 15) = 0$ $\Rightarrow 14x + 14 = 0$ $\Rightarrow x = -1$.Substituting $x = -1$ into $3x + 2y + 5 = 0$: $3(-1) + 2y + 5 = 0$ $\Rightarrow -3 + 2y + 5 = 0$ $\Rightarrow 2y = -2$ $\Rightarrow y = -1$.The point of intersection is $(-1, -1)$.The slope of the line $3x - 5y + 11 = 0$ is $m_1 = \frac{3}{5}$.The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{5}{3}$.The equation of the line passing through $(-1, -1)$ with slope $m_2 = -\frac{5}{3}$ is:$(y - (-1)) = -\frac{5}{3}(x - (-1))$ $\Rightarrow 3(y + 1) = -5(x + 1)$ $\Rightarrow 3y + 3 = -5x - 5$ $\Rightarrow 5x + 3y + 8 = 0$.

The equation of the straight line passing through the point of intersection of $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$ and perpendicular to the line $3x - 5y + 11 = 0$ is

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