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(A) The equations of the lines are:$x-3y+2=0$ ...$(i)$$2x+5y-7=0$ ...(ii)Solving equations $(i)$ and (ii):From $(i)$,$x = 3y-2$. Substituting into (ii

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$2x-3y+1=0$

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(A) The equations of the lines are:$x-3y+2=0$ ...$(i)$$2x+5y-7=0$ ...(ii)Solving equations $(i)$ and (ii):From $(i)$,$x = 3y-2$. Substituting into (ii):$2(3y-2) + 5y - 7 = 0$$6y - 4 + 5y - 7 = 0$$11y = 11 \Rightarrow y = 1$Substituting $y=1$ in $(i)$: $x - 3(1) + 2 = 0 \Rightarrow x = 1$.The point of intersection is $(1, 1)$.The equation of a line perpendicular to $3x+2y+5=0$ is of the form $2x-3y+\lambda=0$.Since this line passes through $(1, 1)$:$2(1) - 3(1) + \lambda = 0$$2 - 3 + \lambda = 0 \Rightarrow \lambda = 1$.Thus,the required equation is $2x-3y+1=0$.

The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$ is:

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