Let the direction cosines of a line be $(l,m,n)$.

They satisfy: $l+m+n=0 \,\,\text{and}\,\, l^2=m^2+n^2$

Since direction cosines also satisfy: $l^2+m^2+n^2=1$

Use the relation $l=-(m+n)$ from $l+m+n=0$

Now substitute into $l^2=m^2+n^2$

Then, $(-(m+n))^2=m^2+n^2$ $m^2+n^2+2mn=m^2+n^2$ $2mn=0$ $mn=0$

So either $m=0$ or $n=0$.

Case 1: $m=0$

Then from $l+m+n=0$, $l+n=0 \Rightarrow l=-n$

Also from $l^2=m^2+n^2$, $l^2=n^2$ which is consistent.

So direction ratios are proportional to: $(1,0,-1)$

Case 2: $n=0$

Then from $l+m+n=0$, $l+m=0 \Rightarrow l=-m$

So direction ratios are proportional to: $(1,-1,0)$

Thus the two lines are along vectors: $\vec{a}=(1,0,-1), \qquad \vec{b}=(1,-1,0)$

Angle $\theta$ between them is given by $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$

Compute dot product: $\vec{a}\cdot\vec{b}=1\cdot 1+0\cdot(-1)+(-1)\cdot 0=1$

Magnitudes: $|\vec{a}|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}$ $|\vec{b}|=\sqrt{1^2+(-1)^2+0^2}=\sqrt{2}$

Hence, $\cos\theta=\frac{1}{\sqrt{2}\cdot\sqrt{2}}=\frac{1}{2}$

Therefore, $\theta=\cos^{-1}\left(\frac12\right)=\frac{\pi}{3}$

So the correct option is: $\boxed{\frac{\pi}{3}}$

Comparison with stored correct answer:

Stored correct answer = C = $\frac{\pi}{3}$, which matches our result.