If $\theta$ is the angle between the lines whose direction cosines are given by $6 m n-2 n l+5 l m=0$ and $3 l+m+5 n=0$, then $\sin \theta=$

A line's direction cosines $(l, m, n)$ satisfy both $6mn - 2nl + 5lm = 0$ and $3l + m + 5n = 0$.

Eliminating $m = -3l - 5n$ via substitution yields the quadratic $l^2 + 3ln + 2n^2 = 0$, which factors as $(l + n)(l + 2n) = 0$.

Case 1: $l = -n$ gives $m = -2n$, producing normalized direction cosines $\left(\frac{-1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$.

Case 2: $l = -2n$ gives $m = n$, producing normalized direction cosines $\left(\frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$.

The cosine of angle $\theta$ between these lines is $\cos \theta = l_1l_2 + m_1m_2 + n_1n_2 = \frac{2}{6} - \frac{2}{6} + \frac{1}{6} = \frac{1}{6}$.

Using $\sin^2 \theta = 1 - \cos^2 \theta$, we find $\sin \theta = \sqrt{\frac{35}{36}} = \frac{\sqrt{35}}{6}$.

Final answer: $\boxed{\text{A}}$