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(B) Let $(h, k)$ be any point on the line $7x - 9y + 10 = 0$. Then, $7h - 9k + 10 = 0$, which implies $h = \frac{9k - 10}{7}$.Perpendicular distance $

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$d_{1} = d_{2}$

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(B) Let $(h, k)$ be any point on the line $7x - 9y + 10 = 0$. Then, $7h - 9k + 10 = 0$, which implies $h = \frac{9k - 10}{7}$.Perpendicular distance $d_{1}$ from $(h, k)$ to $3x + 4y - 5 = 0$ is given by $d_{1} = \frac{|3h + 4k - 5|}{\sqrt{3^{2} + 4^{2}}} = \frac{|3h + 4k - 5|}{5}$.Perpendicular distance $d_{2}$ from $(h, k)$ to $12x + 5y - 7 = 0$ is given by $d_{2} = \frac{|12h + 5k - 7|}{\sqrt{12^{2} + 5^{2}}} = \frac{|12h + 5k - 7|}{13}$.Substituting $h = \frac{9k - 10}{7}$ into the expressions:$3h + 4k - 5 = 3(\frac{9k - 10}{7}) + 4k - 5 = \frac{27k - 30 + 28k - 35}{7} = \frac{55k - 65}{7} = \frac{5(11k - 13)}{7}$.So, $d_{1} = \frac{|5(11k - 13)|}{5 	imes 7} = \frac{|11k - 13|}{7}$.$12h + 5k - 7 = 12(\frac{9k - 10}{7}) + 5k - 7 = \frac{108k - 120 + 35k - 49}{7} = \frac{143k - 169}{7} = \frac{13(11k - 13)}{7}$.So, $d_{2} = \frac{|13(11k - 13)|}{13 	imes 7} = \frac{|11k - 13|}{7}$.Since $d_{1} = \frac{|11k - 13|}{7}$ and $d_{2} = \frac{|11k - 13|}{7}$, we have $d_{1} = d_{2}$.

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point on the line $7x - 9y + 10 = 0$ to the lines $3x + 4y = 5$ and $12x + 5y = 7$, respectively. Then,

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