An angle between the lines whose direction cosines are given by the equations, $$l + 3m + 5n = 0$$ and $$5lm - 2mn + 6nl = 0$$, is:

Let the required lines have direction cosines $$l,\;m,\;n$$. These quantities satisfy both the given relations

$$l + 3m + 5n = 0 \qquad\text{and}\qquad 5lm - 2mn + 6nl = 0.$$

From the linear relation we have

$$l = -\,(3m + 5n).$$

We now substitute this value of $$l$$ into the quadratic equation:

$$5lm - 2mn + 6nl = 0.$$

First we compute each term one by one.

We have $$5lm = 5\bigl(-\,(3m + 5n)\bigr)m = 5\bigl(-3m^2 - 5mn\bigr) = -15m^2 - 25mn.$$

The middle term is already simple:

$$-2mn = -2mn.$$

The last term is

$$6nl = 6n\bigl(-\,(3m + 5n)\bigr) = -18mn - 30n^2.$$

Adding all three expressions and setting the sum equal to zero we get

$$(-15m^2 - 25mn) + (-2mn) + (-18mn - 30n^2) = 0.$$

Simplifying the coefficients of like terms:

$$-15m^2 \;-\;(25 + 2 + 18)mn \;-\;30n^2 = 0,$$

$$-15m^2 - 45mn - 30n^2 = 0.$$

Dividing throughout by $$-15$$ we obtain

$$m^2 + 3mn + 2n^2 = 0.$$

This quadratic in two variables factorises neatly:

$$m^2 + 3mn + 2n^2 = (m + n)(m + 2n) = 0.$$

Hence either

$$m + n = 0 \quad\text{or}\quad m + 2n = 0.$$

We now find the corresponding sets of direction ratios $$\bigl(l,\;m,\;n\bigr).$$

Case 1. If $$m + n = 0,$$ then $$n = -m.$$ Using $$l = -\,(3m + 5n)$$ we get

$$l = -\bigl(3m + 5(-m)\bigr) = -\bigl(3m - 5m\bigr) = -(-2m) = 2m.$$

Thus a suitable set of direction ratios is

$$l : m : n = 2m : m : -m = 2 : 1 : -1.$$

We denote the corresponding vector by

$$\vec v_1 = (2,\;1,\;-1).$$

Case 2. If $$m + 2n = 0,$$ then $$m = -2n.$$ Again, from $$l = -\,(3m + 5n)$$ we find

$$l = -\bigl(3(-2n) + 5n\bigr) = -\bigl(-6n + 5n\bigr) = -(-n) = n.$$

Hence a set of direction ratios is

$$l : m : n = n : -2n : n = 1 : -2 : 1.$$

The corresponding vector is

$$\vec v_2 = (1,\;-2,\;1).$$

We now calculate the angle between the two lines whose direction ratios are given by $$\vec v_1$$ and $$\vec v_2$$. For two vectors $$\vec a = (a_1,a_2,a_3)$$ and $$\vec b = (b_1,b_2,b_3)$$, the cosine of the angle $$\theta$$ between them is defined by the formula

$$\cos\theta = \dfrac{a_1b_1 + a_2b_2 + a_3b_3} {\sqrt{a_1^2 + a_2^2 + a_3^2}\;\sqrt{b_1^2 + b_2^2 + b_3^2}}.$$

Applying this to $$\vec v_1 = (2,1,-1)$$ and $$\vec v_2 = (1,-2,1)$$ we have

$$a_1b_1 + a_2b_2 + a_3b_3 = (2)(1) + (1)(-2) + (-1)(1) = 2 - 2 - 1 = -1.$$

The magnitudes of the vectors are

$$|\vec v_1| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6},$$

$$|\vec v_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.$$

Substituting into the cosine formula we get

$$\cos\theta = \dfrac{-1}{\sqrt{6}\,\sqrt{6}} = -\dfrac{1}{6}.$$

The expression $$\cos\theta = -\dfrac{1}{6}$$ corresponds to an obtuse angle. However, by convention the angle between two lines is taken to be the acute angle, whose cosine is the absolute value of the above result. Therefore

$$\cos\alpha = \left|\cos\theta\right| = \dfrac{1}{6}, \qquad 0^\circ \le \alpha \le 90^\circ.$$

Consequently

$$\alpha = \cos^{-1}\!\left(\dfrac{1}{6}\right).$$

Among the listed choices this corresponds to Option B.

Hence, the correct answer is Option B.