Acute angle between the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$ and $\frac{x-1}{4}=\frac{y+1}{-3}=\frac{z+10}{5}$ is:

$\frac{\pi}{3}$

The correct answer is Option (3) → $\frac{\pi}{3}$

Given lines:

$\displaystyle \frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ and $\displaystyle \frac{x - 1}{4} = \frac{y + 1}{-3} = \frac{z + 10}{5}$

Direction ratios of first line: $(3, 4, 5)$

Direction ratios of second line: $(4, -3, 5)$

Angle between lines is given by:

$\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2}\sqrt{a_2^2 + b_2^2 + c_2^2}}$

$= \frac{|3(4) + 4(-3) + 5(5)|}{\sqrt{3^2 + 4^2 + 5^2} \sqrt{4^2 + (-3)^2 + 5^2}}$

$= \frac{|12 - 12 + 25|}{\sqrt{50} \sqrt{50}} = \frac{25}{50} = \frac{1}{2}$

$\Rightarrow \theta = 60^\circ$

Therefore, the acute angle between the lines is $60^\circ$.