The acute angle between two lines such that the direction cosines $$l, m, n$$, of each of them satisfy the equations $$l + m + n = 0$$ and $$l^2 + m^2 - n^2 = 0$$ is :

The problem requires finding the acute angle between two lines whose direction cosines $$l, m, n$$ satisfy the equations $$l + m + n = 0$$ and $$l^2 + m^2 - n^2 = 0$$. Additionally, direction cosines must satisfy $$l^2 + m^2 + n^2 = 1$$.

Starting with the given equations:

1. $$l + m + n = 0$$

2. $$l^2 + m^2 - n^2 = 0$$

3. $$l^2 + m^2 + n^2 = 1$$ (since $$l, m, n$$ are direction cosines).

From equation 1, express $$n$$ as:

$$n = - (l + m)$$

Substitute this into equation 2:

$$l^2 + m^2 - \left[-(l + m)\right]^2 = 0$$

Simplify the expression:

$$l^2 + m^2 - (l + m)^2 = 0$$

Expand $$(l + m)^2$$:

$$l^2 + m^2 - (l^2 + 2lm + m^2) = 0$$

Distribute the negative sign:

$$l^2 + m^2 - l^2 - 2lm - m^2 = 0$$

Combine like terms:

$$-2lm = 0$$

Thus, $$lm = 0$$, meaning either $$l = 0$$ or $$m = 0$$.

Now consider the two cases.

Case 1: $$l = 0$$

Substitute $$l = 0$$ into equation 1:

$$0 + m + n = 0 \implies m + n = 0 \implies n = -m$$

Use equation 3:

$$(0)^2 + m^2 + (-m)^2 = 1 \implies m^2 + m^2 = 1 \implies 2m^2 = 1 \implies m^2 = \frac{1}{2} \implies m = \pm \frac{1}{\sqrt{2}}$$

Then $$n = -m$$, so if $$m = \frac{1}{\sqrt{2}}$$, then $$n = -\frac{1}{\sqrt{2}}$$, and if $$m = -\frac{1}{\sqrt{2}}$$, then $$n = \frac{1}{\sqrt{2}}$$. Both sets of direction cosines represent the same line (since multiplying by $$-1$$ gives the same direction). Thus, one possible line has direction cosines $$\left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$$.

Case 2: $$m = 0$$

Substitute $$m = 0$$ into equation 1:

$$l + 0 + n = 0 \implies l + n = 0 \implies n = -l$$

Use equation 3:

$$l^2 + (0)^2 + (-l)^2 = 1 \implies l^2 + l^2 = 1 \implies 2l^2 = 1 \implies l^2 = \frac{1}{2} \implies l = \pm \frac{1}{\sqrt{2}}$$

Then $$n = -l$$, so if $$l = \frac{1}{\sqrt{2}}$$, then $$n = -\frac{1}{\sqrt{2}}$$, and if $$l = -\frac{1}{\sqrt{2}}$$, then $$n = \frac{1}{\sqrt{2}}$$. Again, both represent the same line. Thus, another possible line has direction cosines $$\left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$$.

Therefore, the two lines are:

Line 1: $$\left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$$

Line 2: $$\left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$$

The angle $$\theta$$ between two lines with direction cosines $$(l_1, m_1, n_1)$$ and $$(l_2, m_2, n_2)$$ is given by:

$$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$$

Substitute the direction cosines:

$$l_1 l_2 = 0 \cdot \frac{1}{\sqrt{2}} = 0$$

$$m_1 m_2 = \frac{1}{\sqrt{2}} \cdot 0 = 0$$

$$n_1 n_2 = \left(-\frac{1}{\sqrt{2}}\right) \cdot \left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{2}$$

Sum them:

$$0 + 0 + \frac{1}{2} = \frac{1}{2}$$

Thus,

$$\cos \theta = \left| \frac{1}{2} \right| = \frac{1}{2}$$

So $$\theta = 60^\circ$$ (since $$\cos 60^\circ = \frac{1}{2}$$ and $$60^\circ$$ is acute).

Hence, the acute angle between the lines is $$60^\circ$$.

So, the answer is Option C.