If the dr’s of two lines are given by $3lm-4ln+mn=0$ and $l+2m+3n=0$ then the angle between the lines is  

$$\begin{gathered} Given equations are \\ 3lm-4\ln +mn=0\to \left(1\right) \\ and l+2m+3n=0 \\ \Rightarrow l=-2m-3n \\ sub l in \left(1\right) then we get \\ 3m(-2m-3n)-4n(-2m-3n)+mn=0 \\ \Rightarrow -6m^2 -9mn+8mn+12n^2 +mn=0 \\ \Rightarrow -6(m^2-2n^2)=0 \\ \Rightarrow (m+\sqrt{2}n)(m-\sqrt{2}n)=0 \\ \Rightarrow m=-\sqrt{2}n or m=\sqrt{2}n \\ case\left(1\right):if m=-\sqrt{2}n then l=2\sqrt{2}n-3n=(2\sqrt{2}-3)n \\ Now l:m:n=2\sqrt{2}-3:-\sqrt{2}:1 \\ case\left(2\right) :similaly if m=\sqrt{2}n then \\ l:m:n=-2\sqrt{2}-3:\sqrt{2}:1 \end{gathered}$$

$$\begin{gathered} \text{Now D.r's of two lines are }(2\sqrt{2}-3,-\sqrt{2},1),(-2\sqrt{2}-3,\sqrt{2},1) \\ Now a_1{a}_2+b_1{b}_2+c_1{c}_2=0 \\ \therefore lines are perpendicular to each other \end{gathered}$$

$\Rightarrow \text{ Required angle }\theta =\frac{\pi }{2}$