1. Find direction vectors of the lines

For $l_1: \frac{x-2}{3}=\frac{y+1}{-2},\quad z=2$ we can write $x=2+3t,\quad y=-1-2t,\quad z=2.$ So a direction vector of $l_1$ is $\vec d_1=(3,-2,0).$

For $l_2: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ let the common value be $\lambda$. Then $x=1+\lambda,\quad 2y+3=\alpha\lambda\Rightarrow y=\frac{\alpha\lambda-3}{2},\quad z=-5+2\lambda.$ Hence a direction vector of $l_2$ is $\vec d_2=\left(1,\frac{\alpha}{2},2\right).$ A proportional integer form is $\vec d_2=(2,\alpha,4).$

2. Use perpendicularity of $l_1$ and $l_2$

Since the lines are perpendicular, their direction vectors satisfy $\vec d_1\cdot \vec d_2=0.$ Using $(3,-2,0)$ and $(2,\alpha,4)$, $3\cdot 2+(-2)\alpha+0\cdot 4=0$ $6-2\alpha=0$ $\alpha=3.$

Thus direction vector of $l_2$ becomes $\vec d_2=(2,3,4).$

3. Find direction vector of $l_3$

Given $l_3: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}.$ Let the common value be $\mu$. Then $1-x=3\mu\Rightarrow x=1-3\mu,$ $2y-1=-4\mu\Rightarrow y=\frac{1-4\mu}{2},$ $z=4\mu.$ So a direction vector is $\vec d_3=(-3,-2,4).$

4. Angle between $l_2$ and $l_3$

The angle $\theta$ between the lines is the acute angle between their direction vectors.

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

Now, $\vec d_2\cdot \vec d_3=(2)(-3)+(3)(-2)+(4)(4)=-6-6+16=4.$

Magnitudes: $|\vec d_2|=\sqrt{2^2+3^2+4^2}=\sqrt{29},$ $|\vec d_3|=\sqrt{(-3)^2+(-2)^2+4^2}=\sqrt{29}.$

Therefore, $\cos\theta=\frac{4}{29}.$ So, $\theta=\cos^{-1}\left(\frac{4}{29}\right).$

5. Match with options

Option B is $\sec^{-1}\left(\frac{29}{4}\right).$ Since $\sec^{-1}\left(\frac{29}{4}\right)=\cos^{-1}\left(\frac{4}{29}\right),$ this matches our result.

Hence the correct option is B.