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

$\begin{array}{r}{l}_1:\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}-2}{0}\\ l_2:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3/2}{\alpha /2}=\frac{\mathrm{z}+5}{2}\\ l_3:\frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-1/2}{-2}=\frac{\mathrm{z}-0}{4}\\ l_1\perp l_2\Rightarrow 3-\alpha +0=0\Rightarrow \alpha =3\\ \end{array}$

$$\begin{gathered} \text{Let angle between }{l}_2\mathrm{\&}{l}_3 be \theta \\ \begin{array}{l}\cos \theta =\frac{|1\times (-3)+(-2\left)\right(\alpha /2)+2\times 4|}{\sqrt{1+4+\frac{{\alpha }^2}{4}}\sqrt{9+16+4}}\\ \cos \theta =\frac{|-3-\alpha +8|}{\sqrt{5+\frac{{\alpha }^2}{4}}\sqrt{29}},\text{ put }\alpha =3\\ \begin{array}{r}\cos \theta =\frac{2}{\sqrt{\frac{29}{4}}\sqrt{29}}=\frac{4}{29}\\ \theta ={\cos }^{-1}\left(\frac{4}{29}\right)\Rightarrow \theta ={\sec }^{-1}\left(\frac{29}{4}\right)\end{array}\end{array} \end{gathered}$$