Given $l_1:\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$

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

$l_3:\frac{x-1}{-3}=\frac{y-{\displaystyle \frac{1}{2}}}{-2}=\frac{z-0}{4}$

Since $l_1$ is perpendicular to $l_2$, so $\frac{\left|3-\alpha +0\right|}{\sqrt{13}\sqrt{1+\frac{{\alpha }^2}{4}+4}}=0\Rightarrow \alpha =3$

Now for angle between $l_2$ & $l_3$,

$\cos \theta =\frac{\left|1\times \left(-3\right)+(-2)\left({\displaystyle \frac{\alpha }{2}}\right)+2\times 4\right|}{\sqrt{1+\frac{{\alpha }^2}{4}+4}\sqrt{9+4+16}}$

$\cos \theta =\frac{\left|-3-\alpha +8\right|}{\sqrt{5+\frac{{\alpha }^2}{4}}\sqrt{29}}$

Putting $\alpha =3$

$\cos \theta =\frac{2}{\sqrt{\frac{29}{4}}\sqrt{29}}=\frac{4}{29}$

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