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(B) Step $1$: Find the point of intersection of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we get $x = -\frac{1}{17}$ and $y = \

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$119x + 102y = 125$

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(B) Step $1$: Find the point of intersection of $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$. Solving these equations,we get $x = -\frac{1}{17}$ and $y = \frac{22}{17}$. The point is $(-\frac{1}{17}, \frac{22}{17})$.Step $2$: The given line is $6x - 7y + 3 = 0$,which has a slope of $m_1 = \frac{6}{7}$.Step $3$: The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{7}{6}$.Step $4$: Using the point-slope form $y - y_1 = m(x - x_1)$,we have $y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$.Step $5$: Multiplying by $102$,we get $102y - 132 = -119x - 7$,which simplifies to $119x + 102y = 125$.

Find the equation of a line passing through the point of intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ and perpendicular to the line $6x - 7y + 3 = 0$.

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