Let $$\alpha$$ be the angle between the lines whose direction cosines satisfy the equations $$l + m - n = 0$$ and $$l^2 + m^2 - n^2 = 0$$. Then the value of $$\sin^4\alpha + \cos^4\alpha$$ is:

The direction cosines $$(l, m, n)$$ satisfy $$l + m - n = 0$$ and $$l^2 + m^2 - n^2 = 0$$.

From the first equation: $$n = l + m$$. Substituting into the second: $$l^2 + m^2 - (l + m)^2 = 0$$, which gives $$l^2 + m^2 - l^2 - 2lm - m^2 = 0$$, so $$-2lm = 0$$. Therefore either $$l = 0$$ or $$m = 0$$.

Case 1: $$l = 0$$. Then $$n = m$$, giving direction ratios $$(0, 1, 1)$$.

Case 2: $$m = 0$$. Then $$n = l$$, giving direction ratios $$(1, 0, 1)$$.

The angle $$\alpha$$ between these two lines is found using: $$\cos\alpha = \frac{|0 \cdot 1 + 1 \cdot 0 + 1 \cdot 1|}{\sqrt{0^2 + 1^2 + 1^2} \cdot \sqrt{1^2 + 0^2 + 1^2}} = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$$.

So $$\alpha = \frac{\pi}{3}$$, giving $$\sin\alpha = \frac{\sqrt{3}}{2}$$ and $$\cos\alpha = \frac{1}{2}$$.

Therefore: $$\sin^4\alpha + \cos^4\alpha = \left(\frac{\sqrt{3}}{2}\right)^4 + \left(\frac{1}{2}\right)^4 = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8}$$.

The value of $$\sin^4\alpha + \cos^4\alpha$$ is $$\frac{5}{8}$$.