The angle between the lines whose direction cosines satisfy the equations $$l + m + n = 0$$ and $$l^2 = m^2 + n^2$$ is:

Let a set of direction cosines be written as $$(l,m,n)$$. In this problem every admissible triplet must satisfy the two simultaneous relations

$$l+m+n=0\qquad\text{and}\qquad l^{2}=m^{2}+n^{2}.$$

From the first relation we immediately obtain

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

Now we substitute this value of $$l$$ in the second relation. Thus

$$l^{2}=m^{2}+n^{2}\;\;\Longrightarrow\;\;(-(m+n))^{2}=m^{2}+n^{2}.$$

Expanding the left-hand side gives

$$(m+n)^{2}=m^{2}+2mn+n^{2}.$$

So the equality becomes

$$m^{2}+2mn+n^{2}=m^{2}+n^{2}.$$

Subtracting $$m^{2}+n^{2}$$ from both sides leaves

$$2mn=0\;\;\Longrightarrow\;\;mn=0.$$

Hence at least one of $$m$$ or $$n$$ must be zero. We now treat the two possibilities one by one.

Case 1 : $$m=0$$. Putting $$m=0$$ in $$l=-(m+n)$$ gives $$l=-n.$$

Because $$(l,m,n)$$ are direction cosines, they must also satisfy the fundamental relation $$l^{2}+m^{2}+n^{2}=1.$$ Substituting $$m=0$$ and $$l=-n$$ we obtain

$$l^{2}+0+n^{2}=1\;\;\Longrightarrow\;\;2n^{2}=1\;\;\Longrightarrow\;\;n=\pm\dfrac1{\sqrt2},\;l=\mp\dfrac1{\sqrt2}.$$

Thus one line has direction cosines $$\Bigl(\dfrac1{\sqrt2},\,0,\,-\dfrac1{\sqrt2}\Bigr)$$ (changing every sign simultaneously gives the same line).

Case 2 : $$n=0$$. Setting $$n=0$$ in $$l=-(m+n)$$ yields $$l=-m.$$

Again using $$l^{2}+m^{2}+n^{2}=1$$ we get

$$l^{2}+m^{2}+0=1\;\;\Longrightarrow\;\;2m^{2}=1\;\;\Longrightarrow\;\;m=\pm\dfrac1{\sqrt2},\;l=\mp\dfrac1{\sqrt2}.$$

Hence the other line has direction cosines $$\Bigl(-\dfrac1{\sqrt2},\,\dfrac1{\sqrt2},\,0\Bigr).$$

We now know the two distinct lines described by the given conditions. Denote their direction cosines by

$$(l_1,m_1,n_1)=\Bigl(\dfrac1{\sqrt2},\,0,\,-\dfrac1{\sqrt2}\Bigr),\qquad (l_2,m_2,n_2)=\Bigl(-\dfrac1{\sqrt2},\,\dfrac1{\sqrt2},\,0\Bigr).$$

The angle $$\theta$$ between two lines whose direction cosines are $$(l_1,m_1,n_1)$$ and $$(l_2,m_2,n_2)$$ is given by the dot-product formula

$$\cos\theta=l_1l_2+m_1m_2+n_1n_2.$$

Substituting the above numbers, we have

$$\cos\theta=\dfrac1{\sqrt2}\Bigl(-\dfrac1{\sqrt2}\Bigr)+ 0\Bigl(\dfrac1{\sqrt2}\Bigr)+ \Bigl(-\dfrac1{\sqrt2}\Bigr)(0) =-\dfrac12.$$

Therefore

$$\theta=\cos^{-1}\!\Bigl(-\dfrac12\Bigr)=120^\circ.$$

The acute angle between the lines is obtained by subtracting from $$180^\circ$$, so

$$\theta_{\text{acute}}=180^\circ-120^\circ=60^\circ=\dfrac{\pi}{3}\;\text{radians}.$$

Hence, the correct answer is Option C.