If the two lines $$l_1 : \frac{x-2}{3} = \frac{y+1}{-2}, z = 2$$ and $$l_2 : \frac{x-1}{1} = \frac{2y+3}{\alpha} = \frac{z+5}{2}$$ are perpendicular, then an angle between the lines $$l_2$$ and $$l_3 : \frac{1-x}{3} = \frac{2y-1}{-4} = \frac{z}{4}$$ is

Given lines:

$$l_1: \frac{x-2}{3} = \frac{y+1}{-2},\ z = 2$$ — direction vector $$\vec{d_1} = (3, -2, 0)$$

$$l_2: \frac{x-1}{1} = \frac{2y+3}{\alpha} = \frac{z+5}{2}$$

Rewrite $$l_2$$: $$\frac{x-1}{1} = \frac{y + 3/2}{\alpha/2} = \frac{z+5}{2}$$ — direction vector $$\vec{d_2} = (1, \alpha/2, 2)$$

Find $$\alpha$$ using perpendicularity of $$l_1$$ and $$l_2$$.

$$\vec{d_1} \cdot \vec{d_2} = 0$$

$$3(1) + (-2)(\alpha/2) + 0(2) = 0$$

$$3 - \alpha = 0 \Rightarrow \alpha = 3$$

So $$\vec{d_2} = (1, 3/2, 2)$$, or equivalently $$(2, 3, 4)$$.

Find the angle between $$l_2$$ and $$l_3$$.

$$l_3: \frac{1-x}{3} = \frac{2y-1}{-4} = \frac{z}{4}$$

Rewrite: $$\frac{x-1}{-3} = \frac{y - 1/2}{-2} = \frac{z}{4}$$ — direction vector $$\vec{d_3} = (-3, -2, 4)$$, or $$(3, 2, -4)$$.

$$\cos\theta = \frac{|\vec{d_2} \cdot \vec{d_3}|}{|\vec{d_2}||\vec{d_3}|}$$

$$\vec{d_2} \cdot \vec{d_3} = 2(3) + 3(2) + 4(-4) = 6 + 6 - 16 = -4$$

$$|\vec{d_2}| = \sqrt{4 + 9 + 16} = \sqrt{29}$$

$$|\vec{d_3}| = \sqrt{9 + 4 + 16} = \sqrt{29}$$

$$\cos\theta = \frac{4}{29}$$

$$\theta = \cos^{-1}\left(\frac{4}{29}\right) = \sec^{-1}\left(\frac{29}{4}\right)$$

The correct answer is Option B: $$\sec^{-1}\left(\frac{29}{4}\right)$$.