If the direction ratios of two lines are given by $3 l m-4 l n+m n=0$ and $l+2 m+3 n=0$, then the angle between the lines is

Given the equations for direction ratios $(l, m, n)$ are $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$.

From the second equation, $l = -2m - 3n$. Substituting this into the first equation:

$3(-2m - 3n)m - 4(-2m - 3n)n + mn = 0$

$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$

$-6m^2 + 12n^2 = 0$

$m^2 = 2n^2$, which implies $m = \pm \sqrt{2}n$.

Case 1: $m = \sqrt{2}n$. Then $l = -2(\sqrt{2}n) - 3n = -(2\sqrt{2} + 3)n$.
The direction ratios are $(-(2\sqrt{2} + 3), \sqrt{2}, 1)$. Let this be vector $\vec{v_1}$.

Case 2: $m = -\sqrt{2}n$. Then $l = -2(-\sqrt{2}n) - 3n = (2\sqrt{2} - 3)n$.
The direction ratios are $((2\sqrt{2} - 3), -\sqrt{2}, 1)$. Let this be vector $\vec{v_2}$.

The cosine of the angle $\theta$ between the lines is given by $\cos \theta = \dfrac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.

$\vec{v_1} \cdot \vec{v_2} = -(2\sqrt{2} + 3)(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1) = -(8 - 9) - 2 + 1 = 1 - 2 + 1 = 0$.

Since the dot product is $0$, the lines are perpendicular, so $\theta = \dfrac{\pi}{2}$.

Answer: $\dfrac{\pi}{2}$