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(A) The equations of the lines are $x-3y+2=0$ $(i)$ and $2x+5y-7=0$ $(ii)$.Solving equations $(i)$ and $(ii)$ by multiplying $(i)$ by $2$: $2x-6y+4=0$

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$2x-3y+1=0$

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(A) The equations of the lines are $x-3y+2=0$ $(i)$ and $2x+5y-7=0$ $(ii)$.Solving equations $(i)$ and $(ii)$ by multiplying $(i)$ by $2$: $2x-6y+4=0$ $(iii)$.Subtracting $(iii)$ from $(ii)$: $(2x+5y-7) - (2x-6y+4) = 0$ $\Rightarrow 11y - 11 = 0$ $\Rightarrow y=1$.Substituting $y=1$ in $(i)$: $x-3(1)+2=0$ $\Rightarrow x-1=0$ $\Rightarrow x=1$.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 $(1, 1)$,we have $2(1)-3(1)+\lambda=0$ $\Rightarrow 2-3+\lambda=0$ $\Rightarrow \lambda=1$.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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